Informal Notes on Algebra [version 4 Mar 2004 ed.]

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Informal Notes on Algebra R. Boyer

Contents 1 Rings 1.1 Examples and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Integral Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2 2 3 3

2 Ring Homomorphisms 2.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4 4 5

3 Chinese Remainder Theorem 3.1 Field of Quotients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 8

4 Review of Basic Number Theory

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5 Euclidean Domains 5.1 Introduction . . . . . . . . . . . . . . . 5.2 Prime Factorization . . . . . . . . . . 5.3 Gaussian Integers . . . . . . . . . . . . 5.4 Other Examples of Euclidean Domains

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11 11 12 14 17

6 Example of PID that is not a Euclidean Domain 6.1 How to show something is not a Euclidean Domain . . . . . . . . . . . . . . . . . . . 6.2 How to show a ring is a PID . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18 18 18

7 Ring of Polynomials 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Construction of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19 19 21 24

8 Continuation of Polynomials 8.1 Irreducible Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Existence of Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

26 26 26

9 Introduction to Finite Fields

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1

RINGS

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10 Formal Derivatives

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11 Extension Fields

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12 Iterated Field Extensions

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13 Splitting Fields

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14 Galois Group

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15 More Field Extension Results

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16 Discussion Questions about Fields

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17 Galois Correspondence

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18 Galois Theory for Cubic Polynomials 18.1 Solving the Cubic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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19 Galois Theory of Quartic Polynomials

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20 Solvability of Polynomials by Radicals 20.1 Insolvability of the Quintic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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21 Symmetric Functions

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22 Other Descriptions of Galois Theory

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Abstract These are informal notes taken from a variety of sources on basic ring theory and Galois theory.

1 1.1

Rings Examples and Definitions

Definition 1.1. A non-empty set R with two binary operators, written as addition and multiplication, is a ring if it satisfies: (1) R is an abelian group under addition; (2) (Closure) if a, b ∈ R, then ab ∈ R; (3) (Associativity) if a, b, c ∈ R, then (ab)c = a(bc); (4) (Distributivity) if a, b, c ∈ R, then a(b + c) = ab + ac and (b + c)a = ba + ca. Example 1.1. The integers, rational numbers, real numbers, and complex numbers are all rings under the usual operations. Example 1.2. The integers modulo n is a ring.

1

RINGS

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Example 1.3. Matrices whose entries come from any of the previous examples are rings. Definition 1.2. If R1 , R2 , . . . , Rn are all rings, then their direct product R1 × · · · × Rn is a ring under componentwise addition and multiplication. Proposition 1.1. Let a, b ∈ R where R is a ring. Then: (1) a 0 = 0 a = 0; (2) (−a)b = a(−b) = −ab; (3) (−a)(−b) = ab; (4) (n · a)(m · b) = nm · (ab) for all integers m, n. Definition 1.3. Let R be a ring. Then (1) R is a commutative ring if ab = ba for all a, b ∈ R; (2) R is a ring with unity if there is an element denoted by 1 ∈ R such that 1a = a1 = a for all a ∈ R. Definition 1.4. A nonempty subset S of a ring R is a subring of R if for all a, b ∈ S, we have ab ∈ S and a − b ∈ S. Observation 1.1. A subring S of a ring R is a ring. Example 1.4. Consider the ring Z[i] = {a + bi : a, b ∈ Z}, called the ring of Gaussian integers. Note: Z[i] is a subring of the complex numbers C. √ √ Example 1.5. The set Q( 2) = {a + b 2 : a, b ∈ Q} is a ring with the usual operations.

1.2

Integral Domains

Definition 1.5. If a, b ∈ R \ {0} where R is a ring, then we call a and b zero divisors if ab = 0. Proposition 1.2. A nonzero element a ∈ Zn is a zero divisors if and only if a is not relatively prime to n. Corollary 1.1. Zp has no zero divisors if and only if p is prime. Definition 1.6. A ring R is called an integral domain if (1) R is a commutative ring; (2) R has a unity; (3) R has no zero divisors. √ Example 1.6. Z, Zp , Q, R are all integral domains; so are the Gaussian integers Z[i], and Q( 2). Example 1.7. Z × Z is not an integral domains.

1.3

Fields

Definition 1.7. In a ring with unity 1, an element a ∈ R is called a unit if a has a multiplicative inverse in R.

2

RING HOMOMORPHISMS

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Proposition 1.3. Let R be a (commutative) ring with unity 1. Let U (R) = {a ∈ R : a is a unit in R}

(1)

the set of all units in R. Then U (R) is a group under multiplication of R. Proposition 1.4. In Zn , we find U (Zn ) = U (n), the group of all positive integers which are both less than n and relatively prime to n under multiplication. Definition 1.8. A ring R is called a field if (1) R is commutative; (2) R has a unity 1; (3) Every nonzero element in R is a unit. Example 1.8. Q, R, C are all fields. Proposition 1.5. Every field is an integral domain. Proposition 1.6. Every finite integral domain is a field. Corollary 1.2. Zp is a field if and only if p is prime. √ Example 1.9. Q( 2) is a field. Example 1.10. Z3 [i] and Z7 [i] are finite fields. In general, we will show below that Z[i]/ < p >, where p ∈ Z is prime, is a field if and only if p ≡ 3 mod 4. The key condition is whether p can be represented as the sum of two squares. Definition 1.9. A ring R with unity 1 such that every nonzero element a ∈ R is a unit is called a division ring. Definition 1.10. In a ring R, the characteristic of R, denoted char R, is the least positive integer such that n · a = 0 for all a ∈ R. If no such n exists, we say the characteristic of R is 0. Proposition 1.7. Let R be a ring with unity 1. Then 1. char R = 0 if 1 has infinite order under addition; 2. char R = n if 1 has order n under addition. Proposition 1.8. Let D be an integral domain. Then either its characteristic is 0 or is a prime p.

2

Ring Homomorphisms

2.1

Basic Properties

Proposition 2.1. Let φ : R → R0 be a homomorphism between two rings R and R0 . Then: 1. φ(0) = 0; 2. φ(−a) = −φ(a); 3. φ(na) = nφ(a);

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RING HOMOMORPHISMS

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4. φ is injective if and only if Ker(φ) = {0}; 5. φ(a)n = φ(an ), for all n > 0; 6. Ker(φ) is a subring of R. Example 2.1. Consider the polynomial equation 2x3 − 5x2 + 7x − 8 = 0. Claim: this equation has no integer solutions. We argue by contradiction. Let φ : Z → Z3 be the usual map x 7→ x mod 3. Suppose this equation does have an integral solution, say a. Then: 0 = φ(a) = 2a3 − 5a2 + 7a − 8. Note: −5 ≡ 7 ≡ −8 ≡ 1 mod 3. In other words, 2φ(a)3 − φ(a)2 + 7φ(a) − 8 = 2φ(a)3 + φ(a)2 + φ(a) + 1.

(2)

So, if the original equation has a solution a then there must be a solution to 2b 3 + b2 + b + 1 = 0 for some b ∈ Z3 . By exhaustive checking, we find there is no such element b. Observation: Let φ be a homomorphism between two rings, say R1 and R2 . Set a = φ(1). Then it is easy to check that a2 = a. One can use this to show that the only ring homomorphisms of Zn into itself is either the zero homomorphism or the identity.

2.2

Ideals

Definition 2.1. Let R be a ring and I a non-empty subset of R. Then I is an ideal of R if 1. I is a subring of R; 2. For all r ∈ R and x ∈ I, we have rx ∈ I and xr ∈ I. Proposition 2.2. Let φ : R → R0 be a homomorphism between two rings. Then its kernel is an ideal of R. Definition 2.2. Let R be a commutative ring and let a ∈ R. Then the principal ideal generated by a, denoted by < a >, is the set {ra : r ∈ R}. Proposition 2.3. Every ideal of the ring of integers Z is principal. Proposition 2.4. Let R be a commutative ring with unity. Then R is a field if and only if {0} and R are the only ideals in R. Proposition 2.5. Let I be an ideal of R. Then the quotient R/I is a ring with multiplication (a + I)(b + I) = ab + I.

(3)

Theorem 2.1. First Isomorphism Theorem Let φ : R → R 0 be a surjective homomorphism between two rings. Then R/I ∼ = R0 , where I is the kernel of φ. Definition 2.3. A nontrivial proper ideal I of R in a commutative ring R is called a prime ideal if ab ∈ I implies either a ∈ I or b ∈ I for all a, b ∈ R.

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RING HOMOMORPHISMS

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Definition 2.4. A nontrivial proper ideal I of R in a ring R is called a maximal ideal if the only ideals J in R such that I ⊆ J ⊆ R are either I or R. Example 2.2. Let R be the ring of integers. Let U be an ideal of R. CLAIM: U is maximal if and only if U =< p >, where p is prime. Example 2.3. Let R be the ring of all continuous functions on the unit interval [0, 1]. Let M be the ideal of all continuous functions that vanish at the fixed point, say x0 ∈ [0, 1]. CLAIM: M is a maximal ideal of R. Proposition 2.6. Let R be a commutative ring with unity, and let I be an ideal in R. Then 1. I is a prime ideal if and only if R/I is an integral domain 2. I is a maximal ideal if and only if R/I is a field. Proof. (1) Suppose R/I is an integral domain and ab ∈ I. Then (a + I)(b + I) = ab + I = I, which is the zero in the quotient ring. Hence, either a + I or b + I must equal I. In other words, either a ∈ I or b ∈ I, which is the condition for a prime ideal. Next assume I is a prime ideal. Consider (a + I)(b + I) = I in the quotient R/I. Then ab ∈ I so either a ∈ I or b ∈ I. That is, either a + I or b + I must be I. (2) Suppose R/I is a field and J is an ideal of R that properly contains I. Choose b ∈ J \ I. Then b + I must be a non-zero element of R/I. Hence, there is an element c ∈ R so (b + I)(c + I) = 1 + I. Note that 1 − bc ∈ I. On the other hand, bc ∈ J since b ∈ J. We find 1 ∈ J which implies J = R. Suppose I is a maximal ideal and b ∈ R \ I. Then we need to show that b + I has a multiplicative inverse. Consider J = {br + a : r ∈ R, a ∈ I}, which is an ideal that contains both a and I. Then J is an ideal of R that properly contains I. Since I is maximal, we find J = R. In particular, 1 ∈ J so we may find r 0 ∈ R and a0 ∈ A so 1 = br 0 + a0 . Hence (b + I)(r 0 + I) = 1 + I. Corollary 2.1. In a commutative ring R with unity, every maximal ideal is prime. Proof. Let I be a maximal ideal. Then R/I is a field; in particular, it is an integral domain. Example 2.4. Later, we will use maximal ideals of polynomial rings to construct fields. Informally, consider the quotient of A = R[x]/ < x2 + 1 > which will be isomorphic to the field of complex numbers. Let g(x) ∈ R[x]. Then the coset g(x)+ < x2 + 1 > can be represented as a1 x + a0 + < x2 + 1 > by division of polynomials. Further x2 = −1 in A (verify!). We can find the multiplicative inverse, say b1 x + b0 + < x2 + 1 >, of a non-zero element a1 x + a0 + < x2 + 1 >. Set (a1 x + a0 + < x2 + 1 >) (b1 x + b0 + < x2 + 1 >) = 1+ < x2 + 1 > .

(4)

Then a0 b0 − a1 b1 = 1,

a 1 b0 + a 0 b1 .

(5)

To solve for a0 , a1 , we consider two cases: either b0 6= 0 or b1 6= 0. In both cases, we will find that a0 =

−b1 b0 , a1 = 2 . b20 + b21 b0 + b21

Further, we observe that the quotient ring is isomorphic to the field of complex numbers.

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3

CHINESE REMAINDER THEOREM

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Example 2.5. Consider A = Q[x]/ < x2 − 2 >. Then in the quotient x2 = 2. The cosets can be represented as b1 x + b0 + < x2 − 2 >. Again we may check √ that every non-zero coset has a multiplicative inverse. In this calculation, we need to √ use that 2 is irrational. It is interesting to write out the isomorphism between A and the field Q( 2). Example 2.6. Let A = Z[i]/ < 2 − i >. Note: in A we find 2 = i; more precisely 2+ < 2 − i >= i+ < 2 − i >. Hence, every coset representative can be written as a+ < 2 − i > where a ∈ Z. In fact, there are further restrictions since 22 + < 2 − i >= i2 + < 2 − i >= −1+ < 2 − i > in A. Hence, there are only five distinct cosets < 2 − i >, 1+ < 2 − i >, 2+ < 2 − i >, 3+ < 2 − i >, 4+ < 2 − i > (verify!). In fact, one can show that A is isomorphic to Z5 . Definition 2.5. An integral domain R is called a principal ideal domain or a PID if every ideal of R has the form < a >. We saw that the ring Z is a PID. Proposition 2.7. In a PID R every prime ideal is maximal. Proof. Let < p > be a non-zero prime ideal in R. Let I =< m > be any ideal that contains < p >. We must show either I =< p > or I = R. Now p ∈< m > so p = rm for some element r ∈ R. Since < p > is prime and rm ∈< p > either r ∈< p > or m ∈< p >. When m ∈< p >, the ideal I agrees with < p >. When r ∈< p >, write r = sp, where s ∈ R, so p = spm. Since R is an integral domain, we may cancel out the common factor of p to obtain 1 = sm; that is, m is invertible so I = R.

3

Chinese Remainder Theorem

There is a generalization to arbitrary commutative rings with unity of the concept of relatively prime integers m and n. In Z this is equivalent to being able to solve the equation mx + ny = 1. This in turn is equivalent to nZ + mZ = Z as ideals. We shall call two ideals I and J of a ring R comaximal if A + B = R. Recall that the product AB of two ideals is the ideal that consists of all finite sums of the form P j aj bj where a ∈ A and b ∈ B. Moreover, when A and B are principal ideals, say A =< a > and B =< b > we find AB =< ab >. Proposition 3.1. (Chinese Remainder Theorem) Let A1 , A2 , . . . , Ak be ideals in R. Consider the mapping R → R/A1 × R/A2 × · × R/Ak by r 7→ (r + A1 , r + A2 , . . . , r + Ak ) (7) is a ring homomorphism with kernel A1 ∩ A2 ∩ . . . ∩ Ak . If for each i, j ∈ {1, 2, . . . , k} with i 6= j the ideals Ai and Aj are comaximal, then the map is surjective and A1 ∩ A2 ∩ . . . ∩ Ak = A1 A2 . . . Ak . Hence, we have the natural isomorphism R/(A1 A2 . . . Ak ) = R/(A1 ∩ A2 ∩ . . . ∩ Ak ) ∼ = R/A1 × R/A2 × · · · R/Ak .

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Proof. We first show the case when k = 2. Consider the map φ : R → R/A1 × R/A2 defined by φ(r) = (r mod A1 , r mod A2 ).

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CHINESE REMAINDER THEOREM

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This map is a ring homomorphism since r 7→ r mod A1 is just an alternative notation for the natural projection of a ring onto its quotient. Furthermore, the kernel of φ must consist of all elements r ∈ R such that r ∈ A1 and r ∈ A2 ; that is, r ∈ A1 ∩ A2 . Note: all this holds without any restrictions on the ideals A1 and A2 . To complete the proof, we now impose the condition A1 and A2 are also comaximal. We must establish that (1) φ is surjective, and (2) A1 ∩ A2 = A1 A2 . The condition A1 and A2 are comaximal forces A1 +A2 = R. In particular, there must exist elements x ∈ A1 and y ∈ A2 such that x + y = 1. So φ(x) = (0, 1) and φ(y) = (1, 0) (verify). Now let r = (r1 mod A1 , r2 mod A2 ) be an arbitrary element of the product R/A1 × R/A2 . We claim that the element r2 x + r1 y is mapped to r. Consider: φ(r2 x + r1 y)

=

φ(r2 )φ(x) + φ(r1 )φ(y)

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= =

(r2 mod A1 , r2 mod A2 ) (0, 1) + (r1 mod A1 , r1 mod A2 ) (1, 0) (0, r2 mod A2 ) + (r1 mod A1 , 0)

(11) (12)

=

(r1 mod A1 , r2 mod A2 ).

(13)

Hence, the ring homomorphism φ is surjective. It remains to show that A1 ∩A2 = A1 A2 . Now, the ideal A1 A2 is always contained in the intersection A1 ∩ A2 . If A1 and A2 are comaximal and x ∈ A1 and y ∈ A2 are chosen as above, then for any c ∈ A1 ∩ A2 , we have c = c1 = cx + cy ∈ A1 A2 . (14) Hence A1 ∩ A2 ⊂ A1 A2 . The general case follows by induction from the case of two ideals by using A = A 1 and B = A2 A3 · · · Ak once we know that A1 and A2 A3 · · · Ak are comaximal. Corollary 3.1. Suppose that a and b are relatively prime integers. Let α, β ∈ Z. Then there exists an integer x such that x ≡ α(moda), x ≡ β(modb). Proof. Let a and b be relatively prime integers so Z =< a > + < b > so Z/(< a > ∩ < b >) is ring isomorphic to Z/ < a > ⊕Z/ < b >. Hence, given any elements of the rings Z/ < a >, say α+ < a >, and Z/ < b >, say β+ < b >, there must exist an element x ∈ Z that is mapped to (α+ < a >, β+ < b >, by the homomorphism of the proposition. Observation We can also rephrase these results as a structure theorem about the ring Z m . As preparation, suppose m = pq where p and q are distinct primes. Then Zm has ideals I =< p > and J =< q > with zero intersection. Hence Zm is isomorphic to the direct product of Zp and Zq .

3.1

Field of Quotients

Let D be an integral domain. Then there exists a field F consisting of elements written as a/b, where a, b ∈ D with b 6= 0. Moreover, we can identify every element a ∈ D with the element a/1 ∈ F so D becomes a subring of F . Every element has the form a/b = ab−1 where a, b ∈ D with b 6= 0. Any field with these properties is called the field of quotients of D. Further, any two such fields are isomorphic. The explicit construction was outlined in class.

4

4

REVIEW OF BASIC NUMBER THEORY

9

Review of Basic Number Theory

Definition 4.1. We say that c ∈ Z+ is the greatest common divisor of integers a and b if: 1. c|a and c|b, 2. any common divisor of a and b is a divisor of c. Observation 4.1. the greatest common divisor is unique, if it exists. Proposition 4.1.

1. If a, b ∈ Z are not both zero, then their greatest common divisor exists.

2. The greatest common divisor may be written in the form: m0 a + n0 b. Proof : Let S be the set: S = {ax + by : x, y ∈ Z}. Then S must contain a positive integer (verify!) Claim: GCD(a, b) = c = min{ax + by > 0 : x, y ∈ Z}. Now, any common divisor δ of a and b must divide z = ax + by. In particular, δ|c. Next, we must show that c|a and c|b. This will follow by showing that c|z, or c|(ax + by). Now, z = qc + r, where 0 ≤ r < c. That is, r = z − qc = ax + by − qc. Hence, r ∈ S and 0 ≤ r < c. We obtain a contradiction to the minimality of c unless r = 0. We conclude that c|z. In particular, c|a and c|b for proper choices of x and y. Definition 4.2. We call a and b relatively prime if GCD(a, b) = 1. Corollary 4.1. GCD(a, b) = 1 if and only if 1 = ax + by for some choices of x and y. Definition 4.3. Call p > 1 prime if its only positive divisors are 1 and p. Proposition 4.2. If GCD(a, b) = 1 and a|bc, then a|c. Proof : Write 1 = ax + by so c = acx + bcy. Now, a|bcy and a|acx, hence a|c. Corollary 4.2. If p is a prime and divides a product of integers, then it must divide at least one of them. Theorem 4.1. Any positive integer a > 1 is a unique product αk 1 α2 a = pα 1 p2 · · · p k ,

(15)

where p1 > p2 > . . . are prime and each αi > 0. Proof. (Existence) We use induction. The result holds for a = 2. We now assume the result holds for all integers less than a. Now, either a is either prime so the result holds or a = bc, where 1 < b, c < a. By induction b and c are products of primes. Hence, so is a itself. β` β1 β2 αk 1 α2 (Uniqueness) Consider a = pα 1 p2 · · · pk = q1 q2 · · · qk where p1 > p2 > . . . and q1 > q2 > . . . are prime and their exponents are only positive. Claim: k = `, pi = qi and αi = βi for all i. We use induction. The result holds for a = 2. We assume the result for all integers less than a. Since α1 > 0, we find p1 |a so p1 |q1β1 q2β2 · · · qkβ` . (16)

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REVIEW OF BASIC NUMBER THEORY

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In particular, p1 |qi for some i, since p1 is prime. But q1 > qi = p1 . On the other hand, q1 |a implies q1 |pj . As before, p1 ≥ pj ≥ q1 . Hence, q1 ≥ p1 and p1 ≥ q1 implies p1 = q1 . Without loss of generality, assume α1 ≥ b1 . We cancell out one factor of p1 . Then a β1 −1 β2 αk 2 q2 · · · qkβ` . = p1α1 −1 pα 2 · · · pk = p1 p1

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But induction, pi = qi , k = ` and αi = bi for all i. Proposition 4.3. Let p be a prime. Then for any integer a, we have a p ≡ a(modp). Moreover, if p does not divide a, we have ap−1 ≡ 1(modp). Proof. (1) Without loss of generality, we can assume that a is positive. Now, we establish the result using induction on a. The equivalence clearly holds if a = 1. Assume the result for a. We need to establish equivalence for a + 1. Consider: (a + 1)p = 1 + pa + p(p − 1)/2a2 + · · · + ap , by the binomial theorem. All the intermediate terms are divisible by p so are 0 under congruence by p. That is, (a + 1)p ≡ (1 + ap )(modp). By induction, ap ≡ a(modp). We find (a + 1)p ≡ (a + 1)(modp). (2) If p does not divide a, then a is relatively prime to p. In other words, a has a multiplicative inverse modulo p. Multiply both sides of the identity in part (1) by this inverse to obtain (2). Observation: The result that ap−1 ≡ 1(modp) where p is a prime and p does not divide a is called Fermat’s Little Theorem. Proposition 4.4. Chinese Remainder Theorem Suppose that a and b are relatively prime integers. Let α, β ∈ Z. Then there exists an integer x such that x ≡ α(moda),

x ≡ β(modb).

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Proof. We first indicate some reductions. Now it is enough to show that there are integers m, n such that α + ma = β + nb (19) since x ≡ α(moda), x ≡ β(modb) is equivalent to α(moda) ≡ β(modb) which itself is equivalent to the existence of integers m, n such that α + ma = β + nb. To find these integers m, n, it is enough to find other integers s, t such that as + bt = α − β since α + ma = β + nb can be written as α − β = nb − ma. (20) Finally, we can also find solutions to the identity α − β = nb − ma since a and b are relatively prime. That is, we can find integers n0 and m0 so that 1 = n0 b − m0 a. We can simply multiply this last equation by α − β to obtain the desired solution.

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5

EUCLIDEAN DOMAINS

5

11

Euclidean Domains

5.1

Introduction

Definition 5.1. An integral domain R is an Euclidean domain if for every a 6= 0 there is a non-negative integer d(a) such that 1. for all a, b ∈ R, both non-zero, d(a) ≤ d(ab); 2. for any a, b ∈ R, both non-zero, there exist t, r ∈ R such that a = tb + r where either r = 0 or d(r) < d(b) (“division”). Example 5.1. Important examples of Euclidean domains are the ring of integers Z, the ring F [x] of polynomials over a field F , and the ring Z[i] of Gaussians integers. For the Gaussian integers, we will sometimes denote d(a) as N (a) and call it the norm of a; it has the special multiplicative property: N (ab) = N (a)N (b). Proposition 5.1. Let R be a euclidean domain, and let A be an ideal of R. Then A is principal. Proof. If A = {0}, we are done. So assume A = 6 {0}. Choose a0 ∈ A, non-zero, so that d(a0 ) is minimal. Let a ∈ A. By divison, we find a = qa0 + r. By minimality, r = 0. Definition 5.2. Let a 6= 0 and b be elements from a commutative ring R. We say a divides b if there exists c ∈ R so b = ac. Write a | b. It is easy to verify the following: 1. if a|b and b|c then a|c. 2. if a|b and a|c, then a|(b ± c). 3. if a|b, then a|bx for any x ∈ R. Definition 5.3. If a, b ∈ R, then d ∈ R is called the greatest common divisor of a and b if 1. d|a and d|b. 2. Whenever c|a and c|b, then c|d. Proposition 5.2. Let R be a Euclidean domain. Then for any two elements a and b in R have a greatest common divisor. Further, d = xa + yb, for some x, y ∈ R. Proof. Let A be the set of all elements of the form xa + yb, with x, y ∈ R. It is easy to verify that A is an ideal. Hence, A =< d > for some d ∈ R. By construction d = xa + yb. It remains to check that d is the greatest common divisor. Proposition 5.3. Let R be an integral domain with unity. Suppose that for a, b ∈ R, both a|b and b|a are true. Then a = ub where u is some unit in R. Proof. Since a|b, we find b = xa for some x ∈ R. Further, since b|a, we also know a = yb for some y ∈ R. Hence, b = x(yb) = (xy)b. Since R is an integral domain, we can cancel the element b and obtain xy = 1. In particular, y is a unit in R.

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EUCLIDEAN DOMAINS

5.2

12

Prime Factorization

Definition 5.4. Let R be a commutative ring with unity. Two elements a and b are said to be associate if b = ua where u is some unit in R. Proposition 5.4. Let R be a Euclidean domain and a, b ∈ R. If b 6= 0 is not a unit in R, then d(a) < d(ab). Proof. Let A =< a >. Then we find d(a) ≤ d(xa) for any x 6= 0 in R. In particular, d(a) is the smallest d-value for any element in the ideal A. Now if d(ab) = d(a), then the d-value of d(ab) is also minimal. Arguing as before we find that ab is a generator for A. In particular, ab must divide any element of A. So, ab will divide a, that is, a = abx for some x ∈ R. By cancellation, bx = 1 or b is a unit in R. But this contradicts the assumption that b is not a unit. We conclude that d(a) < d(ab). Definition 5.5. In a Euclidean domain R, a non-unit π is called prime or irreducible if whenever π = ab, where a, b ∈ R, then either a or b is a unit in R. Observation 5.1. Some authors make a distinction between prime and irreducible elements. Namely, an element π is prime if < π > is a prime ideal; that is, if π divides ab, then π must divide either a or b. An irreducible element is one given by the above definition. Proposition 5.5. In a PID a non-zero element is prime if and only if it is irreducible. We now show that every element in a Euclidean domain has a unique prime factorization. Proposition 5.6. Let R be a Euclidean domain. Suppose that for a, b, c ∈ R, we have a|bc and GCD(a, b) = 1, then a|c. Proof. Write 1 = xa + yb since 1 is the greatest common divisor. Multiply by c to obtain: c = cax + bcy. Clearly a divides cax and a divides bc by assumption. We conclude that a divides c. Proposition 5.7. If π is a prime element in the Euclidean domain R and π|ab, then π must divide either a or b. Furthermore, if π divides a1 a2 · · · an , then π will divides at least one of the elements a1 , a 2 , . . . , a n . We collect several useful technical results: Proposition 5.8. Let R be a Euclidean domain. Then 1. d(1) is minimal among all d(a), where a ∈ R is non-zero; 2. u ∈ R is a unit if and only if d(u) = d(1); 3. if a and b are associates then d(a) = d(b); 4. for non-zero a, b ∈ R, we have d(a) < d(ab) if and only if b is not a unit in R.

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13

Proof. (1) Suppose a ∈ R is non-zero. Then d(1) ≤ d(1 · a) = d(a). (2) If u is a unit in R, then d(u) ≤ d(uu−1 ) = d(1).

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Hence, d(u) = d(1) for any unit in R. Suppose that a non-zero element u ∈ R satisfies d(u) = d(1). By division, we may find q, r ∈ R so 1 = uq + r

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where either r = 0 or d(r) < d(u). Since d(u) = d(1) is minimal over all d(x) for non-zero x ∈ R, we find that d(r) < d(u) can never hold. Hence, r = 0 and 1 = uq; that is, u is a unit. (3) Since a and b are associates, there must exist a unit u so a = bu. Then u −1 is also a unit and b = au1 . But for non-zero elements x, y ∈ R we always have d(x) ≤ d(xy). In particular, d(b) ≤ d(bu) = d(a) and d(a) ≤ d(au−1 ) ≤ d(b). Hence, d(a) = d(b). (4) Suppose d(a) < d(ab). If b were a unit, then a and ab would be associates and so d(a) = d(ab). Thus, b cannot be a unit. Conversely, we assume d(a) = d(ab). Claim: the ideals < a > and < ab > are equal. The claim follows since the generator z of an ideal is characterized by its minimality among all values d(x) where x is any non-zero element of the ideal. Thus ab will also generate < a >. So we may write a = (ab)c for some c ∈ R. By cancellation in an integral domain, we find 1 = bc. In other words, b is a unit. Proposition 5.9. Let R be a Euclidean domain. Then every element in R is either a unit in R or can be written as the product of finitely many prime elements. Proof. We use induction on the value d(a). If d(a) = d(1), then a is a unit in R as we saw above. The result holds. We assume the result for all elements x such that d(x) < d(a). If a is prime, then we are done. So, suppose that a = bc where neither b nor c are units. We know that d(b) < d(bc) = d(a) and so d(c) < d(a) as well. By the induction hypothesis, we may write both b and c as a product of prime elements. As a consequence, a itself is a product of primes. Theorem 5.1. Let R be a Euclidean domain and a 6= 0 be a non-unit in R. Suppose that a = 0 π1 π2 · · · πn = π10 π20 · · · πm where πi and πj0 are all prime elements in R. Then n = m and each π is 0 an associate of some πj and conversely. Proof. Examine: 0 a = π1 π2 · · · πn = π10 π20 · · · πm .

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0 0 0 π2 · · · πn = u1 π10 π20 · · · πi−1 πi+1 πm .

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0 , say πi0 . Hence, π1 and πi0 are associates; We know that π1 must divide one of primes π10 , π20 , . . . , πm 0 that is, πi = u1 π1 . Next, we may cancel out the factor π1 in the above products to obtain:

We repeat this procedure n times to obtain the equation: 1 = u1 u2 · · · un · z, where z is a certain product of the remaining primes π 0 . We conclude that n ≤ m since the primes π 0 are not units. We reverse the roles of the two factorizations to get m ≤ n. Hence, m = n and each prime πi is an associate of πj0 .

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14

Proposition 5.10. Let ideal A =< a0 > is maximal in a Euclidean domain R if and only if a0 is a prime element. Proof. We first show that if a0 is not prime, then A =< a0 > is not a maximal ideal. Write a0 = bc where b, c ∈ R and neither b nor c are units. Let B =< b >. Then a0 ∈ B so A ⊂ B. To finish we need to verify that A 6= B and B 6= R. If B = R, then 1 ∈ B. So 1 = xb for some x ∈ R. In particular, b is a unit. This violates our assumption on b. If B = A, then b ∈ B = A. So b = xa0 for some x ∈ R. But a0 = bc so b = bcx0 . By cancellation, 1 = cx0 . So c must be a unit. Again, this contradicts our assumption on c. Next, we assume that a0 is a prime element of R. We will show that A is a maximal ideal. Suppose that U is an ideal such that A ⊂ U ⊂ R. Write U =< u0 >. Since a0 ∈ A ⊂ U =< u0 >, we find a0 = xu0 for some x ∈ R. Since a0 is prime, we know that either x or u0 must be a unit and the remaining element is equal to a0 . If u0 is the unit, then U = R. If x is the unit, then u0 = x−1 a0 ∈ A. Hence, U = A.

5.3

Gaussian Integers

For Z[i], we define d(z) = d(a + bi) = a2 + b2 . Proposition 5.11. Z[i] is an euclidean ring. Proof. We just have to verify the divison algorithm: y = tx + r

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where either r = 0 or d(r) < d(x). For our first step we take y to be arbitary but x to be a positive integer. Write: y = a + bi. Then as usual we can find integers u and v so a = un + u1 and b = vn + v1 where u1 and v1 are integers satisfying |u1 | ≤ 21 n and |v1 | ≤ 21 n. Let t = u + vi and r = u1 + v1 i. Then y

= a + bi = un + u1 + (vn + v1 )i

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= (u + vi)n + (u1 + v1 i) = tn + r

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Since d(r) = d(u1 + v1 i) = u21 + v12 ≤ n2 /4 + n2 /4 < n2 = d(n), the result holds for this special case. For the general case, we assume x 6= 0 and y is arbitrary. Now, xx = n is a positive integer. Now, we can apply the above special case of the division algorithm to yx and n to obtain: yx = tn + r (31) where either r = 0 or d(r) < d(n). But n = xx so we have d(yx − txx) < d(n) = d(xx). Next consider the inequalities: d(yx − txx) = d(y − tx)d(x) and d(n) = d(xx), we now have d(y − tx)d(x) < d(x)d(x).

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EUCLIDEAN DOMAINS

15

Since x 6= 0 and d(x) is a positive integer, we may deduce: d(y − tx) < d(x).

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We may write y = tx + r0 where r0 = y − tx. So t and r0 are Gaussian integers. Then either r0 is 0 or d(r0 ) = d(y − tx) < d(x). We conclude that Z[i] is euclidean. Let z ∈ Z[i] whose norm is a prime p in Z. If z = w1 w2 in Z[i] then p = N (w1 )N (w2 ) so one of N (w1 ) or N (w2 ) is ±1 and the other is ±p. Recall an element of Z[i] is a unit if and only if its norm is ±1. Hence, we have Proposition 5.12. If N (z) is ± a prime in Z, then z is irreducible in Z[i]. The converse is false! Let π be a prime element in Z[i] so < π > is a prime ideal in Z[i]. Further, it is easy to check that < π > ∩Z is a prime ideal in Z. Since the norm N (z) = zz is a non-zero integer in < π >, we find < π > ∩Z =< p > for some prime p ∈ Z. Now p ∈< π >. This implies that there must be another gaussian integer, say π 0 so p = ππ 0 in Z[i]. In other words, the prime integer p factors in the larger ring Z[i]. Now N (π)N (π 0 ) = N (p) = p2 . Since π is not a unit, there are two possibilities for the norms: either N (π) = ±p2 or N (π) = ±p. In the former case N (π 0 ) = ±1 hence π 0 must be a unit and p = π (up to associates) is irreducible in Z[i]. In the latter case, N (π) = N (π 0 ) = ±p. Hence π 0 is also irreducible and p = ππ 0 is a product of two irreducibles. In particular, if we write π = a + bi so N (π) = a2 + b2 , we find p = N (π) = a2 + b2 . We sum up these observations: Proposition 5.13. The integer prime p factors in Z[i] into precisely two irreducibles if and only if p = a2 + b2 is the sum of two integer squares. If p = a2 + b2 , the irreducibles are a ± bi. Observation 5.2. Since the square of any integer is congruent to either 0 or 1 modulo 4 (easy to verify!), an odd prime in Z that is the sum of two squares must be congruent to 1 modulo 4. (Here is a quick check: write p = a2 + b2 . If both a2 and b2 are congruent to 1 modulo 4, then their sum must be even.) Hence, a integer prime p in Z congruent to 3 modulo 4 will remain irreducible in the gaussian integers Z[i]. Observation 5.3. The even prime 2 can be written as 12 + 12 with 2 = (1 + i)(1 − i). Proposition 5.14. The prime number p in Z divides an integer of the form n 2 + 1 if and only if p is either 2 or an odd prime congruent to 1 modulo 4. Proof. It is easy to check the case p = 2 since 2 divides 12 + 1. If p is an odd prime, then p divides n2 + 1 is equivalent to n2 = −1 in Zp . This in turn is the same as stating that the residue class of n has order 4 ∈ Z× p , the multiplicative group of non-zero elements of Zp since (−1)2 = 1. Hence p divides an integer of the form n2 + 1 if and only if the multiplicative group Z× p has an element of order 4.

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16

By Lagrange’s Theorem, if Z× p has an element of order 4, 4 must divide the order of the group, that is, 4 divides p − 1 so p is congruent to 1 modulo 4. For the converse, assume p − 1 is divisible by 4. We will show that Z× p has an element of order 4, say n. Then p must divide n2 + 1. To see this, consider n4 ≡ 1 mod p in the group Z× p . Then p must divide n4 − 1 or the product (n2 − 1)(n2 + 1) = (n + 1)(n − 1)(n2 + 1). Since n < p − 1, we find p must divide n2 + 1. Now to the proof that Z× p contains an element of order 4. contains a unique element of order 2. Consider m2 ≡ 1 mod p so m2 − 1 ≡ We first show that Z× p 2 0 mod p. Then p will divide m − 1 = (m − 1)(m + 1). Since p is prime, it must divide either m − 1 or m + 1. If p divides m − 1, then m ≡ 1 mod p; if p divides m + 1, m ≡ −1 mod p. We conclude that −1 is the unique element of order 2. × Next, we show that Z× p contains a subgroup H of order 4. Consider the quotient group Z p /{±1}. × This quotient group must contain a subgroup of order 2, so its preimage H in Z p will be a subgroup of order 4. Now H cannot be isomorphic to Z2 × Z2 for then Z× p would have at least 3 elements of order 2. Hence H is a cyclic subgroup of order 4. Observation 5.4. By the above proposition, if p ≡ 1 mod 4 is an odd prime, then p will divide n2 + 1 for some n ∈ Z. Then p divides (n + i)(n − i) in Z[i]. Now if p is irreducible in Z[i], then p must divide either n + i or n − i. But p is real, so p must divide n + i and its complex conjugate n − i. So p divides (n + i) − (n − i) = 2i which is impossible. Theorem 5.2. 1. The prime p is the sum of two integer squares p = a2 + b2 with a, b ∈ Z if and only if p = 2 or p ≡ 1 mod 4. 2. The irreducible elements in the Gaussian integers Z[i] are: (a) 1 + i with norm 2, (b) the prime p ∈ Z with p ≡ 3 mod 4 with norm p2 ,

(c) a + bi, a − bi, the distinct irreducible factors of p = a2 + b2 = (a + bi)(a − bi) for the prime p ∈ Z with p ≡ 1 mod 4 where both a + bi, a − bi have norm p. Note: the first part of the theorem is the classical result of Fermat on the sum of squares. The Gaussian integers may also be used to characterize all Pythagorean triples. In particular, we wish to determine all integer solutions of x2 + y 2 = z 2 which have no common factor (so-called primitive solutions). If we assume we do have such a solution x, y, z then by considering the equation modulo 4 we find that z must be odd (verify!). We shall show that x + iy has the form uα2 where u is a Gaussian unit and α is a Gaussian integer. Write α itself as m + ni, then we may write: {x, y} = {±(m2 − n2 ), ±2mn},

z = ±(m2 + n2 ).

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It is necessary that m and n are relatively prime and not be both odd (otherwise x, y, and z will have a common factor). Furthermore, it is easy to verify that every primitive Pythagorean triple comes from some choice of m and n as well as a choice of signs. Without loss of generality, we take both m and n to be positive.

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17

Now, assume that π is a Gaussian prime that divides x + iy. To show that x + iy has the form uα 2 it is enough to show that π must divide x + iy an even number e of times. Since (x + iy)(x − iy) = z 2 and π clearly divides z 2 an even number of times, we need only show that π cannot divide x − iy. Now, suppose that π does divide x − iy so it divides both x + iy and its conjugate. In particular π must divide 2x = (x + iy) + (x − iy). It is easy to see that 2x and z are relatively prime (note: z is odd and x, y, and z are relatively prime). Hence, there must exist integers m and n so that 2xm + zn = 1. We conclude that π must divide 1 in Z[i]. This is impossible since π is a prime and cannot be a unit.

5.4

Other Examples of Euclidean Domains

√ The √ argument√to show that the Gaussian integers Z[i] is an Euclidean domain also works for Z[i 2], Z[ 2] and Z[ 3]. A systematic study of the structure of these rings is part of the subject known as algebraic number theory. Here is a brief outline of how to modify the arguments given above. √ Introduce the norm N on R = Z[ 2] by √ (35) N : R → Z, a + b 2 7→ a2 − 2b2 . Then it is straightforward to verify that N (z1 z2 ) = N (z1 )N (z2 ), for all z1 , z2 ∈ R. For the division algorithm, introduce d(z) as |N√(z)|. √ √ Let x = a + b 2 and y = c + d 2 where y 6= 0. Then xy = r + s 2 where r, s ∈ Q. Choose integers √ α, β ∈ Z such that |α − r| ≤ 1/2 and |β − s| ≤ 1/2. Finally, let q = α + β 2. Then µ ¶ x x = qy + y −q . (36) y ´ ³ Then the remainder is z = x − qy or z = y xy − q . We find that µ ¶ x N (z) = N (y)N (y −q (37) y ¡ ¢ = N (y) (α − r)2 − 2b2 (38) ¡ ¢ Now | (α − r)2 − 2b2 | ≤ 2/4 = 1/2 so d(z) ≤ (1/2)d(y) < d(y) as is required. √ ¡ ¢ Identical reasoning with 3 gives | (α − r)2 − 3b2 | ≤ 3/4 so the remainder satisfies d(z) ≤ (3/4)d(y) < d(y). √ √ We can also discuss the units in Z[ 2]. If u ∈√ Z[ 2] is a unit, then N (u)N (u−1 ) = N (1) = 1. Hence N (u) = ±1. In particular, if u = a + b 2, then u is a unit if and only if a2 − 2b2 = ±1. (This√equation is a special case of what is known as Pell’s equation.) One can show √ that the units √ of Z[ 3] have the form ±1 and {±uk : k ∈ Z} where u = 1 + 2. Sometimes 1 + 2 is called a fundamental unit. √ √ A similar result holds for √ Z[ 3]. It has fundamental unit 2 + 3. We can also handle Z[i 2]. Another ring that can be handled with direct methods is Z[ω] where ω = exp(2πi/3), a primitive √ 3 1 cube root of unity so ω = − 2 + 2 i. We can introduce a norm N by: N : Z[ω] → Z,

a + bω 7→ a2 − ab + b2 .

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6

EXAMPLE OF PID THAT IS NOT A EUCLIDEAN DOMAIN

18

It is easy to verify that if a + bω is written in the form u + iv then N (a + bω) = u2 + v 2 . Z[ω] has only six unit elements. We can introduce a division algorithm by imitating yet again the Gaussian integers.

6 6.1

Example of PID that is not a Euclidean Domain How to show something is not a Euclidean Domain

˜ = {0} ∪ {unitsin D}. Then D is a field if and only Let D be an integral domain with unit and let D ˜ if D = D. Proposition 6.1. Let D be a Euclidean domain with “absolute value” function φ : D → Z + . ˜ so that φ(w) is minimal among φ(D \ D). ˜ Then Assume that D is not a field, and choose w ∈ D \ D ˜ for any x ∈ D, there exists z ∈ D such that w|(x − z). Proof. Let w and D be as in the statement, and let x ∈ D. By the division algorithm, we can write ˜ and qw = x − z. x = qw + z, with q, x ∈ D and φ(z) < φ(w). By hypothesis on φ(w), z ∈ D, √ Corollary 6.1. If R = {a + b(1 + −19)/2 : a, b ∈ Z}, then R is not a Euclidean domain. ˜ such that for all x ∈ R, w divides x − z for Proof. It is enought to show that there is√no w ∈ R \ R √ ˜ some z ∈ R. Note first that if a + b(1 + −19)/2 = z ∈ R, then |z|2 = (a + b/2)2 + (b −19/2)2 = a2 + ab + b2 /4 + 19b2 /4 = a2 + ab + 5b2 ∈ Z, so that if z is a unit, a2 + ab + 5b2 = 1 and so b = 0 ˜ = {0, 1, −1}. and a = ±1. Thus R Suppose w has the indicated property. Then w is not a unit and (taking x = 2), w will divide either 1, 2, or 3. Since 2 and 3 are irreducible in √ R (prove this by using the absolute value squared | | 2 ), w = 2, −2, 3, or −3. Now take x = (1 + −19)/2. Then w divides x, x + 1, or x − 1. By the calculation above, |x|2 = 5, |x + 1|2 = 1 + 1 + 5 = 7, and |x − 1|2 = 1 − 1 + 5 = 5. Since neither 5 nor 7 is divisible by 2 or 3, we obtain a contradiction.

6.2

How to show a ring is a PID

Let R be a subring of the complex numbers C, such that |z|2 ∈ Z for all z ∈ R. In particular, the ring R in the previous subsection qualifies. Proposition 6.2. Suppose that for x, y ∈ R with |x| ≥ |y| > 0, either y divides x or else there exist z, w ∈ R with 0 < |xz − yw| < |y|. Then R is a PID. Proof. Let A 6= {0} be an ideal of R. Choose y ∈ A with |y| > 0 minimal (possible since |y| 2 = 1, 2, . . . if y 6= 0) and let x ∈ A. For z, w ∈ R, we have xz − yw ∈ A. Thus either xz − yw = 0 or else |xz − yw| ≥ |y|. By hypothesis, y divides x. Thus A =< y >. √ Corollary 6.2. If R = {a + b(1 + −19)/2 : a, b ∈ Z}, then R is a PID. Proof. Suppose x, y ∈ R and |x| ≥ |y| > 0. If y|x, there is nothing to show, so assume x/y / R. √ ∈ Reducing to lowest terms after rationalizing the denominator, we may write x/y = (a + b −19)/c with a, b, c relatively prime integers and c > 1.

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RING OF POLYNOMIALS

19

Case 1: c ≥ 5. Choose√integers d, e, f, q, √r such that ae + bd + cf = 1, ad − 19be = cq + r, and |r| ≤ c/2. Let z = d + e −19, w = q − f −19. Then √ √ µ ¶ √ √ ¢ a + b −19 ¡ r + −19 x z−w = · d + e −19 − (q − f −19) = . y c c 2

≤ 1 since |r| ≤ c/2 and c ≥ 5. This is non-zero with absolute value r +19 c Case¡ 2: c = 2. √Since¢x/y ∈ / R, we find that a and b must have opposite parity. Let z = 1 and w = (a − 1) + b −19 /2 ∈ R. Then | xy z − w| = 12 < 1. Case 3: c = 3. Since a, b, and c are relatively prime and since √ 0 and 1 are the only squares in Z/ < 3 >, we have a2 + 19b2 ≡ a2 + b2 6 ≡¯ mod 3.¯ Let¯ z = a − b ¯ −19 and choose w ∈ Z so that 2 ¯ ¯ ¯ 2 ¯ a2 + 19b2 = 3w + r with r = 1 or 2. Then ¯ xy z − w¯ = ¯ a +19b − w¯ = | 3r | < 1. 3 Case 4: c = 4. Since our fraction is in lowest terms, a and b are not both even. If they are both 2 2 2 odd, a2 + 19b ≡ a2 + 3b2 ≡ 1¯+ 3 ≡ 4¯ mod so we can ¯ 8, ¯ choose w ∈ Z with a + 19b = 8w + 4. √ 2 2 ¯ ¯ ¯ ¯ − w¯ = 21 < 1. If they are of opposite parity, Let z = a−b 2 −19 ∈ R. Then ¯ xy z − w¯ = ¯ a +19b 8 2 2 a2 + 19b2 ≡ a2 − b2 6 ≡ mod¯ 4, so we¯ can ¯ choose w ∈¯ Z with a + 19b = 4w + r, with r = 1, 2, or 3. √ ¯ ¯ a2 +19b2 ¯ r ¯x Let z = a − b −19. Then ¯ y z − w¯ = ¯ 4 − w¯ = 4 < 1.

7

7.1

Ring of Polynomials Introduction

Definition 7.1. Let R be a ring. A polynomial with coefficients in R and indeterminate x is a finite sum n X f (x) = ai xi = an xn + an−1 xn−1 + · · · a1 x + a0 (40) i=0

where ai ∈ R.

We add polynomials by adding the coefficients of similar powers. Multiplication is given as follows. Let p(x) = a0 +a1 x+· · ·+am xm and q(x) = b0 +b1 x+· · ·+bn xn , then p(x)q(x) = c0 +c1 x+· · ·+ck xk where ct = at b0 + at−1 b1 + · · · + a0 bt . We find that F [x] is a commutative ring with unity. If p(x) = a0 + a1 x + · · · + am xm 6= 0 and am 6= 0, then we call the degree of p(x), written as degf (x), is m. Proposition 7.1. If p(x), q(x) are two non-zero elements of F [x], then the degree of the product p(x)q(x) is the product of their degrees. Proof. Suppose that p(x) = a0 + a1 x + · · · + am xm and q(x) = b0 + b1 x + · · · + bn xn where am 6= 0 and bn 6= 0. Write p(x)q(x) = c0 + c1 x + · · · + ck xk where ct = at b0 + at−1 b1 + · · · + a0 bt . We claim that ct = 0 for t > m + n while cm+n 6= 0. It is easy to check that cm+n = am bn 6= 0 since F is a field. Next we consider ct where t > m+n. Examine the term in its definition ai bj where i+j = t > m+n or j = t − i. If i > m + n then ai = 0 so ai bj = 0. Suppose that i ≤ m + n. Then j = t − i

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20

Proposition 7.2. F [x] is an integral domain. We now show that F [x] is, in fact, an Euclidean domain. The function deg p is defined for all p ∈ F [x] with p 6= 0. It is easy to note that deg p is a non-negative integer and deg p ≤ deg p(x)q(x), for all q(x) 6= 0. We need to establish the “divison algorithm”. Proposition 7.3. Given two polynomials p(x) and q(x) where q 6= 0, then there are polynomials t(x) and r(x) in F [x] such that p(x) = t(x)q(x) + r(x), where either r = 0 or deg r(x) < deg q(x). Proof. If the degree of p is less than the degree of q there is nothing to prove since we can simply take q(x) = 0 and r(x) = p(x). So, we shall assume deg(p) ≥ deg(q). We write: p(x) q(x)

= a 0 + a 1 x + · · · + a m xm = b 0 + b1 x + · · · + b n xn

(41) (42)

where am 6= 0 and bn 6= 0. Let p1 (x) = p(x) − (am /bn )xm−n q(x); then deg p1 < deg p. By induction on the degree of p we find p1 (x) = t1 (x)q(x) + r(x) where either r(x) = 0 or deg(r) < deg(q). In other words, p(x) = p1 (x) + (am /bn )xm−n q(x) = t1 (x)q(x) + (am /bn )xm−n q(x) + r(x). The result now quickly follows. Note: If D is an integral domain which is not a field, then D[x] is not an euclidean domain. The next result is called the “Factor Theorem.” Proposition 7.4. Let F be a field, f (x) be a polynomial in F [x], and α ∈ F . Then α is a zero of f (x) if and only if x − α is a divisor of f (x) in F [x]. Proof. We first consider the case that α is a zero of f (x). Then by the division algorithm, we may write f (x) = q(x)(x − α) + r(x), where r(x) is either 0 or its degree is less than the degree of (x − α) = 1. We verify that r(x) = 0 for otherwise, r(x) is a constant, say c. So, f (x) = q(x) + (x − α) + c which shows that α cannot be a zero of f (x). Contradiction. For the converse, assume that x − α is a divisor of f (x). So, we may write f (x) = q(x)(x − α) which shows that α is a zero of f (x). Proposition 7.5. Let F be a field, f (x) ∈ F [x], and α ∈ F . Then f (α) is the remainder on dividing f (x) by x − α in F [x]. Proof. By division, we write: f (x) = q(x)(x − a) + r(x) where either r(x) is zero or deg(r) < deg(x − a) = 1. So, we find that r(x) must be a constant. The result follows. Proposition 7.6. Let F be a field and f (x) ∈ F [x] be a nonzero polynomial in F [x] of degree n. Then f has at most n zeros in F .

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Proof. We use induction on the degree of f . We use the “factor theorem.” If f has degree 1, then f (x) = a0 + a1 x. But f (x) = a0 (a1 x/a0 + 1). Then f (x) is zero if and only if a1 x/a0 + 1 = 0. We find x = −a0 /a1 . We assume the result for all polynomials of degree strictly less than n. Let f be a polynomial of degree exactly n. If f has no zeros, then result holds. If f has at least one zero, then by the Factor Theorem we find f (x) = f1 (x)(x − α). By induction, f1 has at most n − 1 zeros.

7.2

Irreducible Polynomials

Definition 7.2. A polynomial p(x) is irreducible over F if whenever p(x) = a(x)b(x) then either a(x) or b(x) has degree 0, that is, it is a constant. Note: An ideal in F [x] is maximal if and only if it has the form < p(x) > where p(x) is irreducible. It is difficult to decide if a polynomial is irreducible over a field F in general. Proposition 7.7. Let F be a field. If p(x) is a polynomial of degree 2 or 3, then p(x) is reducible over F if and only if p(x) has a zero in F . Proof. First we suppose that p(x) is reducible over F . Then p(x) = a(x)b(x) where a(x) and b(x) are polynomials over F of degree less than p(x). In particular, one of them, say a(x), has degree 1. So, a(x) = c0 + c1 x where c1 6= 0. Clearly, a(x) has a root and so does p(x). Conversely, if α is a root of p(x) in F , the p(x) will factor as p(x) = (x − α)q(x). Hence, p(x) is reducible. The following result is known as the “rational roots theorem:” Proposition 7.8. Let f (x) = a0 + a1 x + · · · + an xn

(43)

0 = f (a) = an (r/s)n + an−1 (r/s)n−1 + · · · + a1 (r/s) + a0 .

(44)

be a polynomial in Z[x]. Let a be a zero of f (x) in Q. Write a = r/s, where r and s are relatively prime integers. Then r divides a0 and s divides an in Z. Proof. We begin by writing:

By multiplying by sn , we obtain: an rn + an−1 rn−1 s + · · · + a1 rsn−1 + a0 sn = 0.

(45)

By solving for an rn and, separately, a0 sn , we obtain two equations: an r n a0 s n

= −s[an−1 rn−1 + · · · + a1 rsn−2 + a0 sn−1 ] = −r[an rn−1 + · · · + a1 sn−1 ].

Since r and s are relatively prime, we find s divides an and r will divide a0 .

(46) (47)

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Definition 7.3. Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial in Z[x]. Then c = gcd(an , . . . , a0 ) is called the content of f (x), and if c = 1, then f (x) is called a primitive polynomial. Proposition 7.9. Let f (x) and g(x) be two primitive polynomials in Z[x]. Then their product f (x)g(x) is also primitive. Proof. We argue by contradiction. Suppose that f (x)g(x) is not primitive. So there must be a prime number p that divides the content of f (x)g(x). Now there is a natural ring homomorphism φ : Z[x] → Zp [x] that reducing the coefficients of the integral polynomial modulo p. Since p divides every coefficient of f (x)g(x), we must have φ(f (x)g(x)) = φ(f (x)) φ(g(x)) = 0 in Zp [x]. However, Zp [x] is an integral domain, so the product of two elements can be zero only if one of the factors is zero. So, we may take φ(f (x)) = 0 in Zp [x], say. In other words, p must divide all the coefficients of f (x). But this contradicts that the content of f (x) is 1. Proposition 7.10. : Every non-zero polynomial f (x) ∈ Q[x] has a unique factorization f (x) = c(f )f ∗ (x)

(48)

where c(f ) ∈ Q is positive and f ∗ (x) ∈ Z[x] is primitive.

Proof. We write f (x) as a0 /b0 + (a1 /b1 )x + · · · + (an /bn )xn ∈ Q[x]. Let B = b0 b1 · · · bn , so f (x) = (1/B)g(x) where g(x) ∈ Z[x]. Next, define B 0 as the content of g(x) which is positive. Then f (x) = c(f )f ∗ (x) where c(f ) = B 0 /B and f ∗ (x) = (B/B 0 )f (x). Suppose f (x) = dh(x) is a second such factorization, so f ∗ (x) = rh(x), where r = d/c(f ) is a positive rational. Write r = u/v in lowest terms. Then vf ∗ (x) = uh(x) is an equation in Z[x]. Then the coefficients of uh(x) have v as a common divisor. So v must divide all the coefficients of h(x). Since h(x) is primitive, v = 1. Similarly, u = 1. We conclude r = d/c(f ) = u/v = 1. Finally, we have d = c(f ) and f ∗ (x) = h(x). Proposition 7.11. If f (x) ∈ Q[x] factors as f (x) = g(x)h(x), then c(f ) = c(g)c(h)

f ∗ (x) = g ∗ (x)h∗ (x).

and

(49)

Proof. We have: f (x)

= = =

g(x)h(x) ∗

(50) ∗

[c(g)g (x)] [c(h)h (x)] c(g)c(h) g ∗ (x)h∗ (x).

(51) (52)

Since c(g)c(h) is a positive rational number and since the product of two primitive polynomials is primitive and by the uniqueness of the factorization give above, we conclude: c(f ) = c(g)c(h) and f ∗ (x) = g ∗ (x)h∗ (x). Proposition 7.12. (Gauss’s Lemma) Let f (x) be a nonzero polynomial in Z[x]. Then f (x) factors into a product of two polynomials of degrees r and s in Q[x] if and only if f (x) factors into a product of two polynomials of those degrees in Z[x].

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23

Proof. Assume that f (x) = g(x)h(x) in Q[x]. Then f (x) = c(g)c(h)g ∗ (x)h∗ (x) in Q[x], where g ∗ (x), h∗ (x) are primitive polynomials in Z[x]. But c(g)c(h) = c(f ) ∈ Z since f (x) ∈ Z[x]. Hence f (x) = [c(f )g ∗ (x)] h∗ (x) is a factorization in Z[x]. The above proofs can be adapted from integers and their field of quotients to a UFD and its field of quotients. Definition 7.4. An integral domain is a unique factorization domain (UFD) if (1) every element which is not a unit can be factored into primes and (2) this factorization is unique to within order of elements and unit factors. The uniqueness of the content of a polynomial by its choice of being positive is changed to being unique up to multiplication by a unit. The proof that requires the most change is that of Proposition 7.9 since we used a map to Zp . Instead we need to argue directly by writing out the coefficients of the polynomials. As a consequence, we can state: Theorem 7.1. If D is a unique factorization domain, then so is D[x]. Corollary 7.1. If D is a unique factorization domain, then so is D[x1 , x2 , . . . , xn ]. Lemma 7.1. Let D be an integral domain. A non-constant monic polynomial p ∈ D[x] is irreducible if and only if it cannot be factored as a product of monic polynomials of smaller degree. Proof. Let n be the degree of n. Write p(x) = a(x)b(x) where a, b ∈ D[x] are non-constant polynomials of degrees r and s respectively. Then pn = 1 = ar bs ; that is, ar and bs are units in D. Hence p(x) = [bs · a(x)] [ar · b(x)] where bs · a(x) and ar · b(x) are monic polynomials. Example 7.1. It is not true for an arbitrary integral domain D and p ∈ D[x] a monic irreducible polynomial in D[x] that p is also irreducible in F [x] where F is the field of quotients of D. For example, let D = Z[2i] and p(x) = x2 + 1. Then p(x) factors in F [x] as (x − i)(x + i). Furthermore, this example also shows that the integral domain Z[2i] is not a UFD. On the other hand Z[i] is an euclidean domain! There is one final standard result we have not yet discussed: every PID D is a UFD. Here is a sketch of why this is true. Let b ∈ D be non-zero. We need to show that b can be factored uniquely (up to permutation and units) into irreducible elements. To accomplish this, it is sufficient to show that there cannot be an infinite sequence a1 , a2 , a3 , . . . such that each ai is divisible by ai+1 and ai and ai+1 are not associates. Why? Keep factoring a given element until all its factors are irreducible; if this does not happen after finitely many steps then such a sequence results. We assume such an infinite sequence exists. Then there are infinitely many distinct principal ideals < ai > which are nested < a1 >⊂< a2 >⊂ · · · . (53) S∞ We notice that their union i=1 < ai > is itself an ideal of D. In particular, it is principal with a generator, say < a >. But the element a must lie in one of the ideals < ai >, say i = i0 . Then < ai >=< ai0 > for all i ≥ i0 . This contradicts that there are infinitely many distinct ideals. (This is a special case of a more general condition in rings called the ascending chain condition (ACC)).)

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24

It remains to show uniqueness of the factorization. Recall that in a PID each an element p generates a maximal ideal if and only if p is irreducible. Then we find that if an irreducible element p divides ab, then it must divide one of the factors. Hence, if p is irreducible in a PID and p divides the product a1 a2 · · · an , then p must divide one of the factors. With this observation, we can establish uniquenss of the “prime factorization” for PID’s with the same argument as euclidean domains. The following criterion for irreducibility is known as “Eisenstein’s criterion:” Proposition 7.13. Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0

(54)

be a polynomial in Z[x]. Suppose there is a prime number p such that 1. p does not divide an ; 2. p divides ai for all i < n; 3. p2 does not divide a0 . Then f (x) is irreducible over Q.

Proof. Without loss of generality, we may assume that the polynomial f (x) is primitive, for factoring out the greatest common divisor of its coefficients does not alter the hypothesis since p6 |a n . Further, if f (x) factors as a product of two rational polynomials, then by Gauss’s lemma, it factors as the product of two polynomials with integer coefficients. Thus, if we assume that f (x) is reducible, then f (x) = (b0 + b1 x + · · · + br xr ) (c0 + c1 x + · · · + cs xs ), (55) where the b’s and c’s are integers and r, s > 0. For the constant terms, we find a 0 = b0 c0 . Since p|a0 , the prime p must divide either b0 or c0 . Note p cannot divide both since p2 6 |a0 . Now, suppose p|b0 but p 6 |c0 . We must have that p cannot divide all the coefficients b0 , . . . , br since then p would divide all the coefficients of the original polynomial f (x) which cannot hold by assumption. Let bk be the first index not divisible by p so k ≤ r < n. In particular, p will divide all b i for 0 ≤ i < k. But ak = bk c0 bk−1 c1 + bk−2 c2 + · · · + b0 ck . (56) Since p divides ak together with bk−1 , bk−2 , . . . , b0 , we find that p will divide bk c0 . However, the prime p divides neither c0 nor bk . Hence, we obtain a contradiction. In particular, the original polynomial f (x) cannot be factored as proposed.

7.3

Construction of Fields

If p(x) is an irreducible polynomial over a field F , then the quotient ring F [x]/ < p(x) > is a field. We will illustrate this construction explicitly. Let p(x) = x3 − 2 be an irreducible cubic polynomial over the field of rational numbers F . Let A =< p(x) > be the ideal generated by p(x). Then F [x]/A will be a field. We shall indicate what elements look like in the quotient and compute their multiplicative inverses.

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Now, any element in F [x]/ < x3 − 2 > is a coset of the form f (x) + A. Given any polynomial f (x) ∈ F [x], by the division algorithm, we may write f (x) = t(x)(x3 − 2) + r(x) where either r(x) = 0 or deg r(x) < deg(x3 − 2) = 3. In particular, r(x) = a0 + a1 x + a2 x2 , where a0 , a1 , a2 are rational numbers. Hence, in the quotient ring, the coset f (x) + A may be written as: f (x)

= (a0 + A) + (a1 x + A) + (a2 x2 + A) = (a0 + A) + a1 (x + A) + a2 (x2 + A)

(57) (58)

(a0 + A) + a1 (x + A) + a2 (x + A)2

(59)

=

Set t = x + A. Then every element in F [x]/ < x3 − 2 > may be written in the form a 0 + a 1 t + a 2 t2

(60)

Further, the element t satisfies the identity t3 = 2 since t3 − 2 = (x + A)3 − 2 = x3 − 2 + A = A = 0

(61)

in the quotient ring. Note: we can also verify that every coset from F [x]/ < x3 − 2 > has a unique representation as a0 + a1 t + a2 t2 with t3 = 2. It is interesting to verify directly that every non-zero coset has a multiplicative inverse. Let a0 + a1 t + a2 t2 6= 0. Let its inverse be α + βt + γt2 . Then (a0 + a1 t + a2 t2 ) (α + βt + γt2 ) = 1.

(62)

Multiplying out the coefficients and using t3 = 2, we obtain the equations a0 α + 2a2 β + 2a1 γ a1 α + a0 β + 2a2 γ

= =

1 0

(63) (64)

a2 α + a 1 β + a 0 γ

=

0

(65)

This is a linear system for the coefficients α, β, γ. The determinant of the system is given as: a30 + 2a31 + 4a32 − 6a0 a1 a2 6= 0.

(66)

Hence, F [x]/ < x3 −2 > is a field if and only if the only solutions in rational numbers of the equation a30 + 2a31 + 4a32 = 6a0 a1 a2

(67)

is a0 = a1 = a2 = 0. Without loss of generality, it is enough to show that this equation has no integer solutions. Further, we may also assume if a0 , a1 , a2 is an integer solution then these integers are relatively prime. To verify this, write a0 = b0 d, a1 = b1 d, a2 = b2 d where d is the greatest common divisor of a0 , a1 , a2 . Since a30 + 2a31 + 4a32 = 6a0 a1 a2 , we find d3 (b30 + 2b31 + 4b32 ) = d3 (6b0 b1 b2 ). Hence, b30 + 2b31 + 4b32 = 6b0 b1 b2 .

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8

CONTINUATION OF POLYNOMIALS

26

So, we now will assume that a0 , a1 , a2 are relatively prime. Since a30 = 6a0 a1 a2 − (2a31 + 4a32 ), a30 must be even and so a0 itself is even. Write a0 = 2α0 . We obtain 4α03 + a31 + 2a32 = 6α0 a1 a2 .

(69)

Hence, a31 is even and so a1 itself is even. Further, a2 will also be even. At this point, we have shown that a0 , a1 , a2 are all even. But this contradicts the assumption that they are relatively prime. Now, for a solution, a30 must be even. Then a0 itself will be even.

8 8.1

Continuation of Polynomials Irreducible Polynomials

We now give some examples of using the Eisenstein criterion. Recall that if f (x) = a n xn +an−1 xn−1 + · · · + a1 x + a0 is a polynomial in Z[x]. Suppose there is a prime number p such that p does not divide an , p divides ai for all i < n, and p2 does not divide a0 , then f (x) is irreducible over Q. We begin with the specific polynomial x5 − 4x + 2. Clearly it is irreducible over Q using the prime p = 2. More generally, if p(x) is a monic polynomial whose lower order coefficients are all even with a constant term of the form 2 × c where c is odd, then p(x) is irreducible over Q. Definition 8.1. If p is a prime, then the p-th cyclotomic polynomial Φp is given by Φp (x) = (xp − 1)/(x − 1) = xp−1 + xp−2 + · · · + x + 1.

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Proposition 8.1. The p-the cyclotomoic polynomial is irreducible over Q for every prime p. Proof. We begin by noting that a polynomial f (x) is irreducible if and only if f (x + c) is irreducible, for some constant c. In particular, Φp (x) is irreducible if and only if Φp (x + 1) is. But Φp (x + 1) = ((x + 1)p − 1)/x = xp−1 + pxp−2 + p(p − 1)/2 xp−3 + · · · + p, where the coefficients are the binomial coefficients b(p, i). Clearly, the prime p divides all the lower order coefficients and while p 2 does not divide the constant term. The Eisenstein criterion gives a quick proof of the irrationality of the n-th roots of certain integers. Proposition 8.2. If a 6= ±1 is a square-free integer, then xn − a is irreducible over Q for every n ≥ 2. Proof. Since a 6= ±1, there must be a prime p that will divide a. Hence, the Eisenstein criterion applies.

8.2

Existence of Roots

Proposition 8.3. If F is a field and p(x) ∈ F [x] is irreducible, then the quotient F [x]/ < p(x) > is a field containing (an isomorphic copy of ) F and a root θ of p(x).

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INTRODUCTION TO FINITE FIELDS

27

Proof. We know that I =< p(x) > is a maximal ideal in F [x] so the quotient E = F [x]/ < p(x) > must be a field. Note that the map a 7→ a + I is an isomorphism of F to {a + I : a ∈ F } ⊂ E. We identify F with its image so F is a subfield of E. Let θ = x + I ∈ E. Write p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 where ai ∈ F . Then we find p(θ)

= = = = =

(a0 + I) + (a1 + I)θ + · · · + (an + I)θ n

(a0 + I) + (a1 + I)(x + I) + · · · + (an + I)(x + I)n (a0 + I) + (a1 x + I) + · · · + (an xn + I) a 0 + a 1 x + · · · + a n xn + I p(x) + I = I.

(71) (72) (73) (74) (75)

since I =< p(x) >. Since I = 0 + I, I is the zero element in the quotient ring. Hence θ is a root of p(x). Definition 8.2. A polynomial p(x) ∈ F [x] splits over F if it is a product of linear factors; that is, F contains all the roots of p(x). The next result is known as Kronecker’s Theorem: Proposition 8.4. Let f (x) ∈ F [x] where F is a field. Then there is a field E that contains F over which f (x) splits. Proof. We use induction on the degree of the polynomial f (x). If its degree is 1, then it has a root a unique root which lies in the field F itself. We assume the result for all polynomials of degree less than n. Let f (x) have degree exactly n. If f (x) is reducible, then f (x) = p(x)q(x) where p and q both have degrees less than n. Let B be an extension field of F where p(x) splits. Then q(x) ∈ F [x] ⊂ B[x]. So there is a further extension field E of B where q splits again by induction. If f (x) is irreducible, then we saw above that there is an extension field B of F where f has at least one root. In particular, f (x) = p(x)q(x) over B, where both p and q have degrees less than n, that is, f is reducible over B. The result now follows. Later we shall show that the splitting field is unique up to isomorphism.

9

Introduction to Finite Fields

We begin by reviewing the characteristic. Definition 9.1. The prime field of a field F is the intersection of all subfields of F . Proposition 9.1. If F is a field, then its prime field is isomorphic to either Q or Z p where p is a prime.

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Proof. We define the map φ : Z → F by n 7→ n · 1. Then φ is a ring homomorphism. Set I be its kernel. Then Z/I is an integral domain. Hence, I is a prime ideal so I =< p > for some prime or I = {0}. If I = {0}, then φ embeds Z in F so there is an isomorphic copy of Q in F . If I =< p >, then Z/ < p > is finite. So the image is isomorphic to Zp . Definition 9.2. A field has characteristic 0 if its prime field is isomorphic to Q; it has characteristic p if its prime field is isomorphic to Zp . We list some basic results about fields and finite fields. 1. Let f (x), g(x) ∈ F [x]. Then their greatest common divisor gcd(f, g) 6= 1 if and only if there is a field E containing both F and a common root of f (x) and g(x). 2. Let f (x) ∈ F [x]. Then f (x) has no repeated roots if gcd(f, f 0 ) = 1 where f 0 (x) is the formal derivative of f (x). 3. If F has characteristic p, then pa = 0 for all a ∈ F . 4. If F has characteristic p, then (a ± b)p = ap ± bp for all a, b ∈ F . i

5. If F has characteristicd p, then σ : F → F given by a 7→ ap is a field homomorphism. 6. If F has characteristicd p and f (x) ∈ F [x], then (f (x)p ) = f (xp ). 7. Let F be a subfield of a field E. Then we can view E as a vector space over the field F . In particular, if E is a finite field, then E is a finite-dimensional vector space over its prime field Zp for some prime p. Hence E will have pn elements. Proposition 9.2. For every prime p and for every positive integer n, then there exists a finite field with exactly pn elements. Proof. Suppose K is a finite field with q = pn elements. Let K × = K \ {0} is a finite abelian group with q − 1 elements. By Lagrange’s theorem, aq−1 = 1 for all a ∈ K × . In other words, every such element would satisfy xq − x = 0. We now begin the proof. Let g(x) = xq − x ∈ Zp [x]. By Kronecker’s theorem, there is a field E containing Zp over which g(x) splits. We define F = {α ∈ E : g(α) = 0}. (76)

That is, F is the set of roots of g(x). Now g 0 (x) = qx − 1 = −1. Then the greatest common divisor of g and g 0 is 1. Hence, all the roots of g(x) are simple and so there are exactly q roots so |F | = q. To complete the proof, we need to verify that F is closed under addition and multiplication. Since F is finite and an integral domain, it is a field. Let a, b ∈ F so aq = a and bq = b. Now ab is a root of g(x) since (ab)q = aq bq = ab or (ab)q −(ab) = 0. Also, a − b is a root as well since (a − b)q = aq − bq = a − b shows it is a root of g(x). Proposition 9.3. Let F be a finite field with pn elements where p is prime. Then 1. The additive group (F, +) is isomorphic to Zp × Zp × · · · Zp .

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29

2. The multiplicative group (F \ {0}), ∗) is isomorphic to the cyclic group Z pn −1 . Proof. (1) This result follows at once since p · x = 0 for all x ∈ F since F has characteristic p. (2) We begin by observing that F × is an abelian group of order m = pn − 1. Write m = pn − 1 = pe11 · · · pekk , its prime factorization, so the abelian group F × is is isomorphic to the direct product of its Sylow p-subgroups, that is, (77) F∗ ∼ = S(p1 ) × · · · × S(pk )

where S(pi ) is the Sylow subgroup of order pei i . In each Sylow subgroup, choose an element ai of e0

e0

e0

maximal order (so |ai | = pi i ). Then the product element a1 a2 · · · ak ∈ F × has order m0 = p11 · · · pkk since the pi ’s are distinct primes. Next, we saw above that every element x of F ∗ must satisfy xm − 1. Further, since all the roots of xm − 1 are distinct, there is no polynomial xn − 1 = 0, with n < m, which is satisfied by all elements in F ∗ . 0 By construction, xm − 1 = 0 for all elements in S(p1 ) × · · · × S(pk ) since S(pi ) is a p-group. Hence, we find m0 = m.

Examples: 1. We now use the above proof to construct a field F of order 4. Consider the polynomial g(x) = x4 − x over Z2 . Let its roots be 0, 1, α, β. Since {1, α, β} is a group of order 3, it is cyclic so β = α2 . So, the multiplicative structure is determined. Now the abelian group (F, +) has order 4. Now, 2·x = 0 for all x ∈ F since F has characteristic 2. So (F, +) is isomorphic to Z2 × Z2 . The addition is completely determined if we know the sum 1 + α. By elementary properties of group tables, we find 1 + α = α 2 . 2. To construct a field F of order 8 we do not have to use x8 − x. Note that the irreducible cubic p(x) = x3 + x + 1 over Z2 will do. We do this by emphasizing its additive structure F [x]/ < p(x) >. Then the field elements of F may be written as a + bx + cx2 + I

(78)

where a, b, c ∈ Z2 . The additive structure is clear but the explicit mutliplication of elements must be accomplished by using the division algorithm. That is, we first find [(a + bx + cx2 ) + I]

· [(a0 + b0 x + c0 x2 ) + I] = aa0 + (ab0 + a0 b)x + (bb0 + ac0 + a0 c)x2 = =

(79) (80)

+(bc0 + b0 c)x3 + cc0 x4 + I (81) aa + (ab0 + a0 b)x + (bb0 + ac0 + a0 c)x2 + (bc0 + b0 c)(x + 1) + cc0 (x2 + x) + (82) I 0 0 0 0 0 0 0 0 0 0 0 0 2 (aa + bc + b c) + (ab + a b + bc + b c + cc )x + (bb + ac + a c + cc )x (83) +I 0

since x3 = x + 1 and x4 = x2 + x. Observation: Note that even though we have an explicit formula for multiplication it is not clear that the resulting group of non-zero elements is cyclic. Since this group has order 7, though, any non-identity element will be a generator. So, we may express every element of F in the form xj . This is not true in general.

10

FORMAL DERIVATIVES

10

30

Formal Derivatives

Let F be a field with positive characteristic p. Then the derivative of xp is pxp−1 which is zero. So, it is not true for all fields that only constants have zero derivatives. On the other hand, we can show that if f (x) ∈ F [x] where F is a field of characteristic p > 0, then f (x) is a polynomial in x p . Proposition 10.1. For any f (x), g(x) ∈ F [x] and any a ∈ F , we have 1. (f (x) + g(x))0 = f 0 (x) + g 0 (x).

2. (af (x))0 = af 0 (x). 3. (f (x)g(x))0 = f 0 (x)g(x) + f (x)g 0 (x). Proposition 10.2. The polynomial f (x) ∈ F [x] has a multiple root if and only if f (x) and f 0 (x) have a nontrivial common factor. Proof. We begin with an observation. Suppose two polynomials f (x) and g(x) in F [x] have a nontrivial common factor in K[x], for some extension K of F , then these polynomials have a common factor in F [x]. For, if they are relatively prime as elements in F [x], then there are polynomials a(x) and b(x) such that 1 = a(x)f (x) + b(x)g(x). Since this relation will also hold for those elements viewed as elements in K[x], they are relatively prime there as well. Now to the proof. We may assume without loss of generality we may assume that all roots of f (x) will lie in F , otherwise, we extend F to the splitting field of f (x). Suppose f (x) has a multiple root. If f (x) has a multiple root α, then f (x) = (x − α) m q(x) for m > 1. Then f 0 (x) = (x − α)m q 0 (x) + m(x − α)m−1 q(x) = (x − α)r(x). In particular, we find that f and f 0 have a common factor x − α. The converse is easier. Suppose f (x) has no multiple roots, so f (x) = (x − α1 )(x − α2 ) · · · (x − αn ) where the roots are all distinct. Then its derivative is: n X f 0 (x) = (x − α1 ) · · · (x\ − αi ) · · · (x − αn ). (84) i=1

Clearly, no root of f (x) is a root of f 0 (x). So, f and f 0 cannot have a nontrivial common factor, for then they would have a common root. Proposition 10.3. If f (x) ∈ F [x] is irreducible, then 1. If the characteristic of F is 0, then f (x) has no multiple roots. 2. If the characteristic of F is p > 0, then f (x) has a multiple root only if it is of the form f (x) = g(xp ).

Proof. Since f (x) is irreducible, its only factors in F [x] are 1 and f (x). If f (x) has a multiple root, then f and f 0 have a nontrivial common factor, hence f (x) will divide f 0 (x). (Verify: f (x) is an irreducible polynomial so it is a prime element in F [x]. Hence, the greatest common divisor of f (x) and any other polynomial g(x) is either a unit or f (x) itself.) Since the degree of f (x) is less than f 0 (x), we must have f 0 (x) = 0.

11

EXTENSION FIELDS

31 n

Proposition 10.4. If F is a field of characteristic p > 0, then the polynomial x p − x ∈ F [x], has distinct roots. n

n

n

Proof. The derivative of xp − x is pn xp − 1 = −1, since F has characteristic p > 0. Hence, xp − x n and its derivative are relatively prime. By above, we find that xp − x has no multiple roots.

11

Extension Fields

Definition 11.1. The degree of a field K over F is the dimension of K as a vector space over F . Proposition 11.1. Tower Property If L is a finite extension of K and if K is a finite extension of F , then L is a finite extension of F . Moreover,[L : F ] = [L : K] [K : F ]. Proof. Write [L : K] = m with basis w1 , w2 , . . . , wm and [K : F ] = n with basis v1 , v2 , . . . , vn . Claim: the elements B = {vi wj } is a basis for L over K. To verify the claim we shall show that B is a both a spanning set and linear independent. Proposition 11.2. If L is a finite extension of F and K is a subfield of L which contains F , then [K : F ] divides [L : F ]. Proof. This follows easily from the fact that a subspace of a finite dimensional vector space is finite dimensional. Pm Let L have K-basis {w1 , . . . , wm } so any x ∈ L has the form x = j=1 kj wj while K has the F -basis {v1 , . . . , vn }. Then we find à n ! m m X X X fi,j vi wj kj w j = (85) x = j=1

j=1

=

n m X X

i=1

fi,j vi wj .

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j=1 i=1

We find that B = {vi wj } is a F -spanning set for L. To verify linear independence, consider: =

m X n X

fi,j vi wj =

j=1 i=1

à n m X X j=1

i=1

fi,j vi

!

wj .

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Pn Since the vectors {wj } are linearly independent, the coefficients i=1 fi,j vi = 0 for each j. Furthermore, the vectors {vi } are also linearly independent, so the coefficients fi,j themselves are all zero.

12

Iterated Field Extensions

We continue our discussion of fields: F ⊂ K ⊂ L. Recall the result:

13

SPLITTING FIELDS

32

Proposition 12.1. If L is a finite extension of K and if K is a finite extension of F , then L is a finite extension of F . Moreover,[L : F ] = [L : K] [K : F ]. Example Consider the fourth degree polynomial p(x) = x4 − x2 + 9 = (x4 − 6x2 + 9) + 4x2 = (x2 − 3)2 + 4x2 = 0. The equation then becomes [(x2 − 3)/2x]2 = −1. This formula shows that any field which contains a root u of the equation will also contain i = (u2 − 3)/2u since its square is −1. 2 2 In particular, over the field Q(i), the quartic becomes 3− √ reducible and factors as (x − 3 + 2xi)(x −√ 2 2 while the second quadratic has roots i ± 2. 2xi). The quadratic (x − 3 + 2xi) has roots −i ± √ Thus, p(x) has all of its roots in the field Q(i, 2). √ √ If we take F = Q, K = Q( 2), and L = Q(i, 2), then [L : F ] = [L : K] [K : F ] = 4.

13

Splitting Fields

Definition 13.1. If F is a subfield of R, we call E is an extension field of F and write E/F . The dimension of E as a vector space over F is called the degree of E over F and it is denoted by [E : F ]. Say E/F is a finite extension if [E : F ] is finite. An older term for a splitting field is a root field. Proposition 13.1. Let p(x) ∈ F [x] be an irreducible polynomial of degree d. Then E = F [x]/ < p(x) > is an extension field of F of degree d. Proof. Let I =< p(x) > be the ideal in F [x]. Write x + I ∈ E as α. We need to show that S = {1, α, α2 , . . . , αd−1 } is a basis of E over F . By the division algorithm, we sawP before that S is a spanning set for E. d−1 Consider the linear combination i=0 ci αi = 0. That is , α is a root of the polynomial g(x) = Pd−1 i i=0 ci x of degree strictly less than d. The proof will be complete if we can show that the original polynomial p(x) is the polynomial of least degree having α as a root. Suppose g(x) is any polynomial whose degree is strictly smaller than p(x) with f (α) = 0. Then g(x) ∈ I. In particular, p(x) must divide g(x). This is only possible if g(x) = 0. Hence, the elements {1, α, · · · , αd−1 } are linearly independent. Definition 13.2. Let E/F be an extension field and let α1 , . . . , αn ∈ E. Then F (α1 , . . . , αn ) is the smallest subfield of E containing F and α1 , . . . , αn . It is called the field obtained by adjoining α1 , . . . , αn ∈ F . An extension E/F is a simple extension if there exists an element α ∈ F with E = F (α) = {f (α)/g(α) : f (x), g(x) ∈ F [x] and g(α) 6= 0}.

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Definition 13.3. An element α in a field extension L of K is called algebraic over K if there is some polynomial p(x) ∈ K[x] such that p(α) = 0. Otherwise, α is called transcendental over K. The field extension itself is called algebraic if every element of L is algebraic over K. We begin with an easy result. Proposition 13.2. If E is a finite-dimensional extension of F , then E is an algebraic extension field.

13

SPLITTING FIELDS

33

Proof. Let α ∈ E. Let n = dimF (E) = [E : F ]. Then consider the set of elements {1, α, α2 , . . . , αn }. Since there are more than n elements in the this set, it must be linearly dependent. That is, there are field elements fj ∈ F such that n X fj αj = 0. (89) j=0

It follows at once that

Pn

j=0

fj xj is the desired polynomial.

We summarize the basic results in the following: Proposition 13.3. Let E/F be an extension field, and let α ∈ E be algebraic over F . Then 1. There is a monic irreducible polynomial p(x) ∈ F [x] having α as a root. 2. p(x) is the monic polynomial of least degree in F [x] having α as a root, hence is unique. 3. F (α) ∼ = F [x]/ < p(x) > by an isomorphism leaving F pointwise fixed. 4. [F (α) : F ] = deg p. Proof. Choose p(x) as the monic polynomial of least degree in F [x] having α as a root. Note: p(x) exists because α is algebraic. The evaluation map F [x] → F (α) taking f (x) 7→ f (α) is surjective with kernel < p(x) >. By the First Isomorphism Theorem, we find F [x]/ < p(x) > ∼ = F (α). This map will fix the elements of F . Since F (α) is a field, the ideal < p(x) > is maximal so the polynomial itself p(x) is irreducible. We saw above that the degree [F (α) : F ] equals the degree of p(x). Definition 13.4. The polynomial p(x) given above is called the irreducible or minimal polynomial of α over F ; we denote it as mα,F (x) or irrα,F (x) where F may be omitted if it is clear from the context. Proposition 13.4. Suppose K ⊂ L are fields, α ∈ L is algebraic over K, and m α (x) ∈ K[x] is the minimal polynomial for α over K. Then 1. K(α), the subfield of L generated by K and α, is isomorphic to the quotient field K[x]/ < mα (x) >. 2. K(α) is the set of elements of the form k0 + k1 α + · · · + kd−1 αd−1 , where d is the degree of f . 3. dimK (K(α)) = deg(mα . Example 13.1. The polynomial f (x) = x3 − 2x + 2 is irreducible over Q by Eisenstein’s criterion. Every cubic has a real root, so f will have at least one real root, say θ. Then the field Q(θ) ∼ = Q[x]/ < f > consists of elements of the form a + bθ + cθ 2 , where a, b, c ∈ Q. Multiplication is performed by using the distributive law then reducing using the relation θ 3 = 2θ − 2. To find the inverse of an element in Q(θ), we may first compute in Q[x]. Given g(θ) = a + bθ + cθ 2 , there must exist elements r(x), s(x) ∈ Q[x] such that g(x)r(x) + f (x)s(x) = 1, since g and f are relatively prime. (Note: r and s can be computed explicitly by the extended euclidean algorithm.) Hence, r(θ) will be the desired inverse.

14

GALOIS GROUP

34

Let us work out the inverse of 2 + 3θ − θ 2 . Put g(x) = −x2 + 3x + 2. Then: 1 1 (24 + 8x − x2 )g(x) + (35 − 9x)f (x) = 1. 118 118 Hence the desired inverse is

1 118 (24

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+ 8θ − 9θ 2 ).

Example 13.2. Let α be a real root of the irreducible quintic x5 + 2x + 2 over Q. Note: any odd degree polynomial over the reals must have a real root by the Intermediate Value Theorem. In the extension E = field Q(α) of Q, we may write the following elements as a 0 + a1 α + a2 α2 + a3 α3 + a4 α4 where (α3 + 2)(α3 + 3α), 1/α, and (α + 2)/(α2 + 3).

14

Galois Group

We begin by noting that the set of all automorphisms of a field K forms a group under composition. Usually, we are interested in a field K which is a field extension of a subfield F . Then we restrict our focus on the automorphisms T of K which leave the elements pointwise fixed, that is, T f = f for any element f ∈ F . Example 14.1. The field C of complex numbers is an extension field of the reals. There are only two automorphisms of C that leave the reals fixed; namely, the identity and the map a + bi 7→ a − bi. √ Example 14.2. Let E = Q( 2) be the extension field of F = Q. √ √ Again there are only two automorphisms of E that fix F : the identity and the map a+b 2 7→ a−b 2. Definition 14.1. The automorphism group of a field K over a subfield F is the group of those automorphisms of K which leave every element of F fixed. Definition 14.2. We call two elements u and v of a field E which is an extension of a field F conjugate if u and v are both roots of the same irreducible polynomial p(x) over the field F . Example 14.3. Let ω be a primitive cube root of 1. √ √ Then ω 2 and ω 2 2 are conjugate over Q. Proposition 14.1. Any automorphism T of a finite extension E of F maps each element α of E onto a conjugate T α of α over F . Proof. We need to show that both α and T α satisfy the same irreducible polynomial over F . Since the extension E is finite over F , we know that α is algebraic over F . Further, we may let p(x) be the unique monic irreducible polynomial p(x) ∈ F [x] of minimal degree such that p(α) = 0. Since T is an automorphism of E that fixes the elements of F , it is straightforward to verify that T (p(α)) = p(T (α)) = 0. √ Example 14.4. Let E = Q( 2, i) which has degree 4 over Q.

14

GALOIS GROUP

35

√ √ Further [E : K] = 2 where K = Q(i). Of course, i and −i are conjugate as well as 2 and − √2. Hence any √ automorphism T of E that fixes the elements of Q must map i to either ±i and 2 to either ± 2. In particular, we find there are exactly four such automorphisms and the resulting group is isomorphic to Z2 × Z2 . Definition 14.3. If E = F (α1 , . . . , αn ) is the splitting field of a polynomial f (x) = (x − α1 ) · · · (x − αn ), then the automorphism group of E over F is called the Galois group of the equation f (x) = 0 or the Galois group of the field E over F . Discussion Suppose E is the splitting field of a polynomial f (x) with distinct roots α 1 , . . . , αn . Then any element T of the Galois group effects a permutation σ of the roots by T (α i ). Further, any element in the splitting field E can be written as a polynomial in the roots with coefficients from F . In particular, for w ∈ E, w = h(α1 , . . . , αn ) where h ∈ F [x1 , . . . , xn ]. Since T leaves these coefficients of h fixed, we find T (h(α1 , . . . , αn )) = h(T α1 , . . . , T αn ) = h(ασ(1) , . . . , ασ(n) ).

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Hence, we have a natural homomorphism from the Galois group into the group of permutations of the roots. We summarize this discussion in the following: Proposition 14.2. 1. Let f (x) ∈ F [x] with exactly k distinct roots α1 , . . . , αn in its splitting field E. Then every automorphism T in the Galois group determines a unique permutation of the roots. 2. The Galois group of any polynomial is isomorphic to a group of permutations of its distinct roots. 3. The Galois group of a polynomial of degree n has order dividing n!. Note: Later we shall give a numerical test for polynomials over Q that determines when the Galois group is a subgroup of the alternating group. Example 14.5. Consider the polynomial x3 − 1 over Q. √ √ Then its roots are 1, ω = exp[i2π/3] = (−1 + 3i)/2, and ω 2 = exp[i4π/3] = (−1 − √3i)/2. Then its splitting field E is generated by ω only and its degree over Q is 2. In fact, E = Q( 3i). Example 14.6. Consider the polynomial x3 − 2 over Q, which is irreducible over Q, unlike the previous example. √ √ √ √ + 3i)/2 is a primitive cube-root of √ 1. Let Then its roots are 3 2, ω 3 2, and ω 2 3 2 where ω = (−1 √ E be its splitting field. Then E must contain both 3 2 and ω individually. Hence, E = Q( 3 2, ω) which has degree 6 over Q by the Tower Property: √ √ √ 3 3 3 (92) [E : Q] = [Q( 2, ω) : Q( 2)] [Q( 2) : Q]. Now, AutQ (E) is isomorphic to S3 , the full group of permutations of the roots. To see this, let T be √ the automorphism of order 3 that takes ω 7→ ω 2 but leaves 3 2 fixed; while S is the automorphism

14

GALOIS GROUP

36

of order 2 that is the restriction to E of automorphism of C over R given by complex conjugation. It is straightforward to verify that the automorphisms S and T do not commute, so the Galois group AutQ (E) is indeed the unique non-abelian group of order 6, S3 . Note that each of the fields Q(α) where α is one of the roots of x3 − 2, is a cubic (that is, degree 3) extension of Q, since Q(α) = Q(α, α2 ) (verify!). Furthermore, it is the fixed field of the automorphism subgroup of order 2 that fixes α while interchanges the other two roots. The other fixed field is √ Q( 3i) which is fixed by the automorphism T of order 3. Example 14.7. Galois group of x4 − 3 over Q. The polynomial is irreducible over Q by the Eisenstein criteria with prime √ p = 3. It has four distinct roots r, ir, −r, and −ir where r is the unique positive fourth root of 3, 4 3. The splitting field E may be given as Q(r, ir, −r, −ir) = Q(r, i). Now [Q(r, i) : Q] = [Q(r, i) : Q(r)][Q(r) : Q]. But [Q(r) : Q] = 4 and [Q(r, i) : Q(r)] = 2 so the degree of E over Q is 8. Let 1, r, r 2 , r3 , i, ir, ir 2 , ir3 be a linear basis for E over Q. Now, any automorphism T from the Galois group will be uniquely determined by its values T (r) and T (i). Recall that T x must be a conjugate of x. Of course the conjugates of r are r, ir, −r, −ir (with minimal polynomial x4 − 3) while the conjugates of i are i, −i (with minimal polynomial x2 + 1). Hence, there are 8 possible elements of the Galois group. Let T be the automorphism determined by r 7→ ir and i 7→ i. Let S be the automorphism given by r 7→ r and i 7→ −i. Then T 4 = I and S 2 = I. The elements of the group G are given by: {I, T, T 2 , T 3 , S, T S, T 2 S, T 3 S}. It is isomorphic to the dihedral group D4 since T S = S 3 T . Note: S 3 (r) = −ir and S 3 (i) = i. Here is a group table for the dihedral group D4 :

r i

I r i

T ir i

T2 −r i

T3 −ir i

S r −i

TS ir −i

T 2S −r −i

T 3S −ir −i

Here is a lattice of the subgroups.

% B1 = {I, S}

A1 = {I, T 2 , S, T 2 S} ↑ B2 = {I, T 2 S} -

% -

G = {I, T, T 2 , T 3 , S, T S, T S 2 , T 3 S} ↑ A2 = {I, T, T 2 , T 3 } ↑ % B3 = {I, T 2 } ↑ % {I}

A3 = {I, T 2 , T S, T 3 S} ↑ B4 = {I, T S} B5 = {I, T 3 S} %

(93) We can also set up a correspondence between the subgroups of the Galois group and subfields of the splitting field.

14

GALOIS GROUP

F1 = Q(r) &

37

. F2 = Q(ir) ↓ G1 = Q(r 2 )

E = Q(r, i) ↓ F3 = Q(r 2 , i) . ↑ G2 = Q(i) & ↓ Q .

& &

& F4 = Q((1 − i)r) ↓ G3 = Q(ir 2 )

F5 = Q((1 + i)r) .

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.

Note that the dimensions dimFj (E) = 2 for j = 1, 2, 3, 4, 5 and dimGj (Fk ) = 2,if Gj ≤ Fk , or dimGj (E) = 4. This will follow from 17.2 given below. Below, we show by direct calculation that Fix(Bj ) = Fj while Fix(Ak ) = Gk ; that is, the Galois correspondence between subgroups of AutQ (E) and subfields of the splitting field E reverses containment. We give several examples of how to verify the lattice of subfields. Let A2 = {I, T, T 2 , T 3 }. Among the eight basis elements of Q(r, i) over Q, the subgroup A2 fixes only the element i. Hence, its fixed field is Fix(A2 ) = Q(i). Let B1 = {I, S}. To find its fixed field requires a bit more work. Recall S(r) = r and S(i) = −i. Now the extension Q(r) over Q has degree 4 since the minimal polynomial min r,Q (x) = x4 − 3. Furthermore, the other roots of minr,Q (x) are also fixed by S (verify!). Since S 6= I, Fix(B1 ) = Q(r). Let B4 = {I, T S}. We will show that Fix(B4 ) = Q(r(1 + i). By direct computation, we see that r + ir = r(1 + i) is fixed by T S. We need to find the degree of the algebraic element γ = r + ir. Since γ ∈ E, the possibilities are 1, 2, 4, or 8. Since AutQ (E) is not cyclic, no element of E has degree 8. Since γ ∈ / Q, γ does not have degree 1. Of course r 4 = 3 and (1 + i)4 = −4. Hence, γ 4 = −12 so minγ,Q(x) (x) = x4 + 12 and γ has degree 4. Let B5 = {I, T 3 S}. Then the calculation of Fix(B5 ) = Q(r(1 − i) is very similar to the B4 case. Definition 14.4. A polynomial p(x) of degree n is separable over the field F if it has n distinct roots in some splitting field E ⊃ F . Otherwise, we call p(x) inseparable. A finite extension E of F is called separable over F if every element of E satisfies a separable polynomial equation. Comment: Any irreducible polynomial p(x) ∈ F [x] over a field F of characteristic 0 is separable. Examples: √ √ 1. Describe the Galois group of Q( 2, 3) over Q as a subgroup of the permutations of the roots of (x2 − 2)(x2 − 3) ∈ Q[x]. We begin with:

√ √ √ √ √ √ [Q( 2, 3) : Q] = [Q( 2, 3) : Q( 2)] [Q( 2) : Q]

so there are four elements in the Galois group. There are: α1 = identity, then √ √ √ √ √ √ √ √ α2 : 2 → − 2, − 2 → 2, 3 → 3, − 3 → − 3 √ √ √ √ √ √ √ √ α3 : 2 → 2, − 2 → − 2, 3 → 3, − 3 → 3 √ √ √ √ √ √ √ √ α4 : 2 → − 2, − 2 → 2, 3 → − 3, − 3 → 3

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(96) (97) (98)

14

GALOIS GROUP

38

√ √ √ √ If we number the roots 2, − 2, 3, − 3 as 1, 2, 3, 4, then the above automorphisms correspond to certain permutations in S4 . They are: α2 ↔ (1, 2),

α3 ↔ (3, 4),

α4 ↔ (1, 2)(3, 4)

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There √ to each subgroup of the Galois group: Fix(α 2 ) = √ are intermediate√subfields that correspond Q( 3), Fix(α3 ) = Q( 2), and Fix(α3 ) = Q( 6). These are fairly easy to verify! 2. Consider the Galois group of Q(ζ) over Q where ζ is a primitive 8-th root of unity. Now [Q(ζ) : Q] is 4 because ζ is a root of the irreducible polynomial x4 + 1. Note: ζ 4 = −1 since ζ 8 = 1. The other primitive 8-th roots of unity are given by the odd powers of ζ, that is, ζ 3 , ζ 5 , ζ 7 . These are exactly the roots of the polynomial x4 + 1. The Galois group has three non-trivial automorphisms: α1 : ζ → ζ 3 ,

α2 : ζ → ζ 5 ,

α3 : ζ → ζ 7

(100)

These automorphisms correspond to permutations in S4 given by: α1 ↔ (1, 2)(3, 4),

α2 ↔ (1, 3)(2, 4),

α3 ↔ (1, 4)(2, 3).

(101)

Of course this group is isomorphic to Z2 × Z2 .

Another way to get the Galois group is to start with ζ = √12 + √12 i. We can verify that √ √ √ 2 ∈ Q(ζ) and i ∈ Q(ζ). But [Q( 2, i) : Q] = 4. Hence, Q( 2, i) = Q(ζ). √ But Q( 2, i) is the splitting field of the polynomial (x2 − 2)(x2 + 1) over Q. √ √ The roots are 2, − 2, i, −i. Hence, we find: √ √ √ √ α : 2 → − 2, − 2 → 2, i → −i, −i → i (102) √ √ √ √ β : 2 → − 2, − 2 → 2, i → i, −i → −i (103) √ √ √ √ γ : 2 → 2, − 2 → − 2, i → −i, −i → i (104) Hence, α ⇐⇒ α1 ,

β ⇐⇒ α2 ,

γ ⇐⇒ α3 .

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3. Consider E the splitting field of the eigth-degree polynomial x8 − 2. Let ζ be a primitive 8-th degree of unity and θ the unique positive 8-th root of 2. Then E = Q(θ, ζ). √ √ We first show that Q(ζ) = Q(i, 2). To verify this, note that we may take ζ = 22 (1+i). Then √ the other roots are ±1, ±i, and 22 (±1 ± i) (where the choice of signs is made independently). √ Since θ 4 = 2, we find the splitting field E = Q(θ, i). Next, we verify that [E : Q] = 16. Now the subfield Q(θ) is of degree 8 over Q since x 8 − 2 is irreducible by the Eisenstein criterion and all its roots are powers of θ. Further, because

14

GALOIS GROUP

39

Q(θ) ⊂ R and i ∈ / Q(θ) but i2 ∈ Q(θ), [Q(θ, i) : Q(θ)] = 2. Hence, [Q(θ, i) : Q] = [Q(θ, i) : Q(θ)] [Q(θ) : Q] = 16. As an aside, note that θ and ζ are algebraically dependent: √ θ4 = 2 = ζ + ζ 7

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The automorphisms in the Galois group AutQ (E) are determined by their action on the field generators θ and i. The possibilities are θ 7→ ζ a θ, a = 0, 1, . . . , 7;

i 7→ ±i.

(107)

We recall that an automorphism must map a root θ onto another root of the irreducible polynomial x8 − 2. Similarly, i must go to ±i, which are the roots of x2 + 1. The Galois group will have two generators σ and τ given by: σ

:

τ

:

θ 7→ ζθ, i 7→ i;

θ 7→ θ, i 7→ −i.

(108) (109)

We need to find the action of σ and τ on the field element ζ. We observe: √ √ √ 2 2 1 +i = (1 + i) 2 ζ = 2 2 2 1 4 (1 + i)θ = 2

(110)

since θ is the positive 8-th root of 2. We compute for the automorphism σ: µ ¶ 1 1 4 σ(ζ) = σ ( (1 + i)θ = (1 + i)σ(θ 4 ) 2 2 1 (1 + i)ζ 4 θ4 = ζ 4 ζ = ζ 5 = −ζ. = 2

(112)

(111)

(113)

and for the automorphism τ : τ (ζ)

= =

µ

1 (1 + i)θ 4 2

¶

1 (1 − i)τ (θ 4 ) 2 √ 1 1 (1 − i)θ 4 = (1 − i) 2 = ζ 7 . 2 2

τ

=

(114) (115)

We can verify the relations between the two generators of the Galois group: σ 8 = τ 2 = Id, στ = τ σ 3 .

(116)

4. Suppose that the cubic f (x) = x3 + ax2 + bx + c is an irreducible cubic over Q and has only one real root. Then its Galois group is isomorphic to S3 .

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MORE FIELD EXTENSION RESULTS

40

Proof. Let u1 be the real root of f (x) while u2 and u3 the complex roots. Note: u2 and u3 must be complex conjugates of one another. Of course, the Galois group must be Q(u1 , u2 , u3 ) a subgroup of S3 . Since [Q(u1 , u2 , u3 ) : Q] divides 6, u2 and u3 are roots of an irreducible quadratic so [Q(u2 , u3 ) : Q] = 2, and u1 ∈ / Q(u2 , u3 ) so [Q(u1 ) : Q(u2 , u3 )] = 3. 5. Later we shall show that the Galois group for xn − 1 over Q is always abelian.

15

More Field Extension Results

We will be following the text by Fred Goodman here. Proposition 15.1. Let B and B 0 be fields, σ : B → B 0 be an isomorphism, and f (x) ∈ B[x] be irreducible. Suppose α is a root of f (x) in some extension field, say L, of B, and α 0 be a root of σ(f (x)) in an extension field L0 of B 0 . Then there exists an isomorphism ψ : B(α) → B 0 (α0 ) such that ψ(b) = σ(b) for all b ∈ B and ψ(α) = α0 . Proof. As usual, σ extends to give an isomorphism of the polynomial rings, σ : B[x] → B 0 [x]. In turn, this induces a field isomorphism σ ˜ : B[x]/ < f >→ B 0 [x]/ < f 0 > such that σ ˜ (b+ < f >) = σ(b)+ < f 0 >, ∀b ∈ B.

(117)

But B[x] ∼ = B 0 [x]/ < f 0 >. = B[x]/ < f > and B 0 [x] ∼ ˜≤L ˜ with an isomorphism σ : B → tildeB. Let p(x) ∈ B[x] Proposition 15.2. Let B ≤ L and B ˜ ˜ for p˜(x). Then and p˜(x) ∈ B[x] such that p˜(x) = σ(p(x)). Let L be a splitting field of p(x) and L ˜ there exists a isomorphism τ : L → L such that τ (b) = σ(b) for all b ∈ B. Proof. We use induction and Proposition 15.1. ˜ = L. ˜ If dimB (L) = 1, then p(x) factors over B and so does p˜(x). Hence B = L and B ˜ ˜ ˜ ˜ Induction Hypothesis: Let B ≤ M ≤ L and B ≤ M ≤ L; σ ˜ : M → M be an isomorphism ˜ such that extending σ; and dimM (L) < n = dimB (L), then there exists an isomorphism τ : L → L τ (m) = σ ˜ (m), for all m ∈ M . L ↑ M ↑ B

τ

−→ σ ˜

−→ σ

−→

˜ L ↑ ˜ M ↑ ˜ B

(118)

Since dimB (L) = n > 1, at least one irreducible factor, say p1 (x), of p(x) has degree > 1, we find that σ(p1 (x)) is an irreducible factor of p˜(x). ˜ be roots of p1 (x) and p˜(x), respectively. Let α ∈ L and α ˜∈L ˜ α) by α 7→ α By Proposition 15.1, there exists an isomorphism σ ˜ : B(α) → B(˜ ˜ extending σ.

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41

By induction, we are done! We apply the induction hypothesis to: L ↑ M

τ

−→ σ ˜

−→

˜ L ↑ ˜ M

(119)

˜ that extends σ We get an isomorphism τ : L → L ˜ ; in particular, τ extends σ as well. ˜ are both splitting fields for p(x). Then there Proposition 15.3. Let p(x) ∈ B[x]. Suppose L and L ˜ such that τ (b) = b for all b ∈ B. exists an isomorphism τ : L → L Proof. Take σ to be the identity map of B → B and apply Proposition 15.2. The technique of proof of Proposition 15.2 allows us to count automorphisms. Let E be the splitting field of f (x) ∈ B[x]. Of course, we know that any isomorphism φ : B → B 0 can be extended to an isomorphism σ : E → E 0 between the splitting fields of f (x) and f 0 (x) = φ(f (x)) ∈ B 0 [x]. We use induction on dimB (E) that the number of such extensions is bounded by dimB (E), with equality if f (x) is separable over B. If dimB (E) = 1, then E = B and the number of extensions is 1. The result holds. If dim B (E) > 1, then f (x) has at least one irreducible factor, say p(x) of degree greater than one with corresponding irreducible factor p0 (x) of f 0 (x). Let α be a fixed root of p(x). If σ is any extension of φ to E, then σ restricted to the subfield B(α) of E is an isomorphism τ of B(α) with some subfield of E 0 . Moreover, this isomoprhism is completely determined by the value τ (α). Note: τ (α) must be a root, say β of p0 (x). Hence, we have the familiar diagram: E ↑ B(α) ↑ B

σ

−→ τ˜

−→ φ

−→

E0 ↑ 0 B (β) ↑

(120)

B0

Conversely, for any root β of p0 (x), there are extensions τ and φ realizing such a diagram. Hence to count the number of extensions σ of φ, we need only count the number of such diagrams. Now, the number of extensions φ to an isomorphism τ is equal to the number of distinct roots β of p0 (x). Since the degree of both p(x) and p0 (x) equals dimB (B(α)), we find that the number of extensions of φ to a τ is at most dimB (B(α)) with equality if all the roots of p(x) are distinct. Since E is the splitting field of f (x) over B(α), E 0 is the splitting field of f 0 (x) over B 0 (x) and dimB(α) (E) < dimB (E), we find by induction that the number of extensions of τ to σ is bounded above by dimB(α) (E) with equality if f (x) has distinct roots. From dimB (E) = dimB(α) (E) dimB (B(α)), it follows that the number of extensions of φ to σ is bounded by dimB (E). Equality holds if p(x) and f (x) have distinct roots, which is equivalent to f (x) having distinct roots since p(x) is a factor of f (x). Of course, if the special case that B = B 0 and φ is the identity map, we have f (x) = f 0 (x) and E = E 0 so the isomorphisms of E to E 0 restricting to φ on B are the automorphisms of E fixing B. Hence we have shown the following:

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MORE FIELD EXTENSION RESULTS

42

Proposition 15.4. Let E be the splitting field of f (x) ∈ B[x]. Then |Aut B (E)| ≤ dimB (E) with equality if f (x) is separable. Note: equality holds for any irreducible polynomial over a field of characteristics zero. Proposition 15.5. Finite fields are unique up to isomorphism. Proof. We constructed a finite field of order pn , where p is a prime, as the splitting field of the n polynomial xp − x ∈ Zp [x]. Question: let E1 and E2 be splitting fields of monic irreducible polynomials p1 (x) and p2 (x) over Q. Then E1 can equal E2 if the roots of p1 (x) are listed as α1 , α2 , . . . , αn , T is an automorphism of E1 that fix Q, and the roots of p2 (x), say β1 , . . . , βn differ from those of T (α1 ), . . . , T (αn ) by rationals; that is, T (α1 ) − α1 ∈ Q. Proposition 15.6. Let f (x) ∈ B[x] with splitting field L; let p(x) be an irreducible factor of f (x) with roots α and β of p(x) in L. Then there exists an automorphism σ ∈ AutB (L) such that σ(α) = β. Proof. Apply Proposition 15.3 to: L ↑ B(α)

σ

−→ −→

L ↑ B(β).

(121)

Proposition 15.7. Let p(x) ∈ B[x] and let L be its splitting field. Assume B ≤ M ≤ L and B ≤ M 0 ≤ L. Let σ : M → M 0 be an isomorphism that leaves B pointwise fixed. Then σ extends to a B-automorphism of L or σ ∈ AutB (L). Proof. Apply Proposition 15.2 to: L ↑ M

−→ σ

−→

L ↑ ˜. M

(122)

˜ [x] so σ(p(x)) = p(x). Treat p(x) as in M [x] and p(x) ∈ M Proposition 15.8. Let p(x) ∈ B[x] and let L be its splitting field. Let B ≤ M ≤ L. Then: (a) There is a bijection between the left cosets of AutM (L) in AutB (L) onto IsomB (M, L). (b) |IsomB (M, L)| = [AutB (L) : AutM (L)]. Proof. By Proposition 15.7, the map τ → τ |M is a surjection of AutB (L) onto IsomB (M, L). Claim: τ1 |M = τ2 |M if and only if τ1 , τ2 lie in the same left coset of AutM (L) in AutB (L). To verify the Claim, we see that τ1 |M = τ2 |M if and only if τ1 τ2−1 |M is the identity on M ; that is, τ1 τ2−1 lies in AutM (L). Of course, this is equivalent that τ1 and τ2 lie in the same coset in AutB (L)/AutM (L). We can use this Proposition to find the minimal polynomial of elements in an extension field. Let √ us re-examine example 14.7. There we let γ = r(1 + i) where r = 4 3. Claim: minγ,Q(x) (x) = x4 + 12 so γ has degree 4 over Q.

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MORE FIELD EXTENSION RESULTS

43

Now for any conjugate of γ over Q, there is an automorphism that takes γ to that conjugate. Hence, to find the minimal polynomial we need to find the distinct values of σ(γ) for σ ∈ Aut Q (E). But the automorphisms that give distinct images for γ are just the representatives of the ledt cosets of AutQ(γ) (E) in the Galois group AutQ (E). There are {I, T, T 2 , T 3 }. So the conjugates are r + ir, ir − r, −r − ir, and −ir + r. Hence min (x) = [(x − (r + ir) (x − (r − ir)] [(x − (−r + ir) (x − (−r − ir)] = x4 − 12.

γ,Q(x)

(123)

Proposition 15.9. Let B ≤ L and H ≤ AutB (L). Then: (a) Fix(H) ≤ L; (b) H ≤ AutFix(H) (L); (c) B ≤ Fix(AutB (L)). Proposition 15.10. (a) If H1 ≤ H2 ≤ AutB (L), then Fix(H1 ) ≤ Fix(H2 ). (b) If K1 ≤ K2 ≤ L, then AutK2 (L) ≤ AutK1 (L). Proposition 15.11. Fix(AutFix(H) (L)) = Fix(H). Proof. Let f ∈ Fix(AutFix(H) (L)). Then α(f ) = f for all α ∈ AutFix(H) (L). But H ≤ AutFix(H) (L) so f ∈ Fix(H). On the other hand, assume α(f ) = f , for all α ∈ H. But β ∈ AutFix(H) (L) must fix any element of Fix(H), hence β(f ) = f . The result follows. Proposition 15.12. Let L be a splitting field for a separable polynomial f (x) ∈ B[x]. Then Fix(AutB (L)) = B. Proof. Let β1 , β2 , . . . , βr be the distinct roots of f (x) in L. Consider the sequence of fields: M0 = B ≤ · · · ≤ Mj = B[β1 , . . . , βj ] ≤ . . . ≤ Mr = B[β1 , . . . , βr ] = L.

(124)

Of course, B ≤ Fix(AutB (L)). Claim: If a ∈ Mj for j ≥ 1, then a ∈ Mj−1 . The result follows from the claim! To start, suppose a ∈ Mj . If Mj−1 = Mj , we are done! So, let ` be the dimension [Mj−1 : Mj ]. Assume ` > 1. Then there is a Mj−1 -basis form Mj of the form: {1, βj , βj2 , . . . , βj`−1 }.

(125)

a = m0 + m1 βj + · · · + m`−1 βj`−1 ,

(126)

Then the element a can be written as

where m0 , . . . , m`−1 ∈ Mj−1 . Let q(x) be the minimal polynomial of βj over Mj−1 of degree `. Then q(x) must divide the original polynomial f (x) since βj is a root of f (x). Moreover, since f (x) is a separable polynomial so must q(x). Hence, the roots of q(x), say {α1 = βj , α2 , . . . , α` } (127) are all distinct and lie in L.

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MORE FIELD EXTENSION RESULTS

44

Recall that for each index s, there is an automorphism σs ∈ AutMj−1 (L) ≤ AutB (L) such that σs (α1 ) = αs . Apply σs to the expansion of a relative to the Mj−1 -basis over Mj to obtain: ¢ ¡ (128) σs (a) = σs m0 + m1 βj + · · · + m`−1 βj`−1 a

=

m0 + m1 αs + · · · + m`−1 αs`−1 ,

(129)

since σs leaves the elements of Mj−1 fixed and σs (a) = a by assumption (a ∈ Fix(AutB (L))). Consider the polynomial (m0 − a) + m1 x + · · · + m`−1 x`−1 of degree ≤ ` − 1. It has at least ` roots in L, namely α1 , α2 , . . . , α` . Hence, this polynomial must be identically zero. This implies that α ∈ Mj−1 . Proposition 15.13. Suppose B ≤ L, dimB (L) < ∞, and Fix(AutB (L)) = B. (a) For all β ∈ L, the element β is algebraic and separable over B; its minimial polynomial min β,B (x) splits in L[x]. (b) For all β ∈ L, let β = β1 , . . . , βn be a list of the distinct elements of the set {σ(β) : σ ∈ AutB (L)}. Then (x − β1 ) · · · (x − βn ) is the minimial polynomial for β over B. (c) L is the splitting field of a separable polynomial in B[x]. Proof. Let Qr β ∈ L. We list the distinct elements of {σ(β) : σ ∈ AutB (L)} as β = β1 , . . . , βr . Next, let g(x) = j=1 (x − βj ) ∈ L[x]. Further, σ(g(x)) = g(x) for all σ ∈ AutB (L). Hence, the coefficients of g(x) are in the fixed field Fix(AutB (L)) = B, by assumption. Let p(x) = minβ,B (x). Then p(x) must divide g(x) since β is a root of g(x). On the other hand, every root of g(x) has the form σ(β). Hence, it must be a root of p(x) as well because any conjugate of a root of a minimal polynomial is still a root. Since the roots of g(x) are all simple, we find that g(x) must divide p(x). We find g(x) = p(x) because both are monic. Hence, the minimal polynomial splits over B. Finally, to verify part (c), note that dimB (L) < ∞. So L is generated over B by finitely many algebraic elements, say α1 , . . . , αs . By part (a), L is the splitting field of the polynomial f = f1 f2 · · · fs , where fi = minαi ,B (x). Proposition 15.14. For a finite-dimensional field extension L of B, the following are equivalent: (a) Fix(AutB (L)) = B; (b) The extension is separable, and for all α ∈ L, the minimal polynomial min α,B splits into linear factors over L; (c) L is the splitting field of a separable polynomial in B[x]. Definition 15.1. Call L a Galois extension of B if it satisfies any of the above equivalent conditions. Proposition 15.15. If B ≤ L is a finite-dimensional Galois extension and B ≤ M ≤ L is an intermediate field, then M ≤ L is a Galois extension. Proof. Since L is a splitting field of a separable polynomial over B, it is a splitting field of the same polynomial considered as an element of M [x]. Proposition 15.16. If B ≤ L is a finite-dimensional Galois extension, then dim B (L) = |AutB (L)|.

16

DISCUSSION QUESTIONS ABOUT FIELDS

45

Proof. The result follows immediately if B = L. We use the induction hypothesis that if B ≤ M ≤ L is an intermediate field and dim M (L) < dimB (L), then | dimM (L)| = |AutM (L)|. Let p(x) ∈ B[x] be the minimal polynomial of α over B. Since L is Galois over B, p(x) is separable and splits over L, by Proposition 15.14. If φ ∈ IsomB (B(α, L), then φ(α) is a root of p(x), and φ is uniquely determined by the value φ(α). Hence: deg(p) = |IsomB (B(α), L)| = [AutB (L) : AutB(α) (L)].

(130)

By the induction hypothesis applied to B(α), we find |AutB(α) (L)| = dimB(α) (L) is finite. Hence, the group AutB (L) is also finite, and |AutB (L)|

= deg(p) ∗ |AutB(α) (L)| = dimB (B(α)) ∗ dimB(α) (L) = dimB (L).

(131) (132)

Proposition 15.17. Let B ≤ L be a finite-dimensional Galois extension and M be an intermediate field, then |IsomB (M, L)| = dimB (M ). (133) Proof. Consider the equations: |IsomB (M, L)| = [AutB (L) : AutM (L)] =

16

dimB (L) = dimB (M ). dimM (L)

(134)

Discussion Questions about Fields

Here are some points of discussion to get more familiar with field theory before continuing! √ √ 1. The fields Q( 3) and Q( 5) are NOT isomorphic. Hint: Mention splitting fields. √ 2. (a) Find all the conjugates of 3 + 2 over Q. p √ (b) Find all the conjugates of 1 + 2 over Q. p p √ √ √ (c) Answer: ± 1 + ± 2 Find all the conjugates of 1 + 2 over Q( 2). Answer: p √ 1+± 2 3. Mark the following true or false: √ √ (a) The fields Q( 2) and Q( 3) are isomorphic.

(b) For all α, β ∈ E, there is always an automorphism of E mapping α onto β.

(c) For α, β algebraic over a field F , there is always an isomorphism of F (α) onto F (β).

(d) For α, β algebraic and conjugate over a field F , there is always an isomorphism of F (α) onto F (β).

16

DISCUSSION QUESTIONS ABOUT FIELDS

46

(e) Every automorphism of every field E leaves fixed every element of the prime field. (f) Every automorphism of every field E leaves fixed an infinite number of elements of E. (g) Every automorphism of every field E leaves fixed at least two elements of E. (h) Every automorphism of every field E of characteristic zero leaves fixed an infinite number of elements of E. (i) The set of all elements of a field E left fixed by a single automorphism of E is a subfield of E. (j) For fields B ≤ M ≤ E, we have Gal(K/E) ≤ Gal(K/F ). AutM (E) ≤ AutB (E). 4. Let Φp (x) be the cyclotomic polynomial (xp−1 − 1)/(x − 1) = xp−1 + xp−2 + · · · + x + 1. We saw it is irreducible over Q for every prime p. Let ζ be one of its roots. (a) The roots of Φp (x) are ζ, ζ 2 , . . . , ζ p−1 . They are all distinct. (b) The Galois group AutQ (Q(ζ)) is abelian of order p − 1. (c) The fixed field is exactly Q.

5. The only automorphism of the reals is the identity. To see this, consider an automorphism of R. Then it maps squares to squares. In particular, it will map positive elements to positive elements. Further, it will preserve the order relation 1, not all the roots of f (x) can lie in F . In particular, f (x) has at least one irreducible factor, say p(x) of degree d with 1 < d ≤ m. Let u be a root of p(x) in N and let p0 (x) be the factor of f 0 (x) corresponding to p(x) under the isomorphism Φ. The splitting field N then must contain a root u0 of p0 (x). By the above result we find that Φ can be extended to an isomorphism Φ∗ such that Φ∗ [F (u)] = F 0 (u0 ),

Φ∗ (u) = u0 ,

p(u) = 0,

p0 (u0 ) = 0.

(186)

Since N is generated over F by the roots of f (x), then N is certaining generated over the larger field F (u) by these roots. So, N is a splitting field of f (x) over F (u), with degree m/d. Similarly, N 0 is a splitting field of f 0 (x) over F 0 (u0 ). Since m/d < m, the induction assumption implies that the isomorphism Φ ∗ can be extended from F (u) to N . Proposition 22.4. If the polynomial f (x) in the above Proposition is separable, the Φ can be extended to N in exactly m = [N : F ] ways. Proof The result is found by a similar induction argument. Any extension Ψ of the given isomorphism Φ must map the root u to one of the roots u 0 of p0 (x). Hence, every possible extension of Φ is given by one of our constructions. Since f (x) is separable, the irreducible factor p(x) of degree d will have exactly d distinct roots u 0 . These d choices give exactly d choices for Φ∗ . By induction, each such map Φ∗ can be extended to N in exactly m/d ways. We find that there are d(m/d) extensions. The result is established. Comments: 1. If the two splitting fields N and N 0 are both extensions of the same base field F , and Φ is the identity map of F onto itself, then we find that N is isomorphic to N 0 2. On the other hand, if f (x) is separable and if N and N 0 are identical, then the identity map can be extended in m different ways to an automorphism of N . 3. These extensions are exactly the automorphisms of the Galois group of N over F . Hence, the order of the Galois group equals [N : F ]. 4. Let G be the Galois group of the splitting field N of a separable polynomial over F . Let K be the field of all elements fixed pointwise by the elements of the Galois group. (It is easy to verify that K is indeed a field and that F ⊂ K.)

Now, every automorphism in G is an extension to N of the identity automorphism of K. Since N is a splitting field over K, there are exactly [N : K] such extensons.

22

OTHER DESCRIPTIONS OF GALOIS THEORY

59

On the other hand, the Galois group has order [N : F ]. We conclude [N : F ] = [N : K]. In particular, the fixed field is precisely F itself. Definition 22.1. A finite extension E of a field F is called normal over F if every polynomial p(x) irreducible over F which has one root in E has all its roots in E. Proposition 22.5. A finite extension of F is normal over F if and only if it is the splitting field of some polynomial over F . Proof If N is normal over F , choose any element u of N not in F and find the irreducible polynomial p(x) = 0 satisfied by u. By the definition of normality, all the roots of p(x) must lie in N . Hence, N will contain the splitting field M of p(x). If there are elements of N not in M , one of these elements v will satisfy an irreducible equation g(x) = 0 and M is contained in the larger splitting field of p(x)g(x). We can continue this process. Eventually, since the degree of N over F is finite, one of the successive splitting fields must be the entire field N . The converse is a little more difficult. Suppose there is a polynomial p(x) irreducible over F which has one but not all its roots in N . Let w be a root of p(x) in N , and adjoin to N another root w 0 which is not in N . Then the simple extension F (w) is isomorphic to F (w 0 ) by the map Φ takes w to w 0 . The field N is a splitting field for f (x) over F (w). On the other hand, N 0 = N 0 (x) is generated by the roots of f (x) over F (w 0 ). Hence, it is a splitting field for f (x) over F (w 0 ). We saw above that this map can be exteneded to an isomorphism Φ∗ of N to N 0 . Since Φ∗ will leave the elements of the base field F fixed, these isomorphic fields N and N 0 must have the same degree over F . But we assumed that N 0 = N 0 (w0 ) is a proper extension of N so its degree over F is larger than that of N . Contradiction.