Informal Notes on Algebra

Informal Notes on Algebra R. Boyer Contents 1 Rings 2 1.1 Examples and Definitions................................. 2 1.2 Integral Domains...................................... 3 1.3 Fields............................................ 3 2 Ring Homomorphisms 4 2.1 Basic Properties...................................... 4 2.2 Ideals............................................ 5 3 Chinese Remainder Theorem 7 3.1 Field of Quotients..................................... 8 4 Review of Basic Number Theory 9 5 Euclidean Domains 11 5.1 Introduction......................................... 11 5.2 Prime Factorization.................................... 12 5.3 Gaussian Integers...................................... 14 5.4 Other Examples of Euclidean Domains.......................... 17 6 Example of PID that is not a Euclidean Domain 18 6.1 How to show something is not a Euclidean Domain................... 18 6.2 How to show a ring is a PID................................ 18 7 Ring of Polynomials 19 7.1 Introduction......................................... 19 7.2 Irreducible Polynomials.................................. 21 7.3 Construction of Fields................................... 24 8 Continuation of Polynomials 26 8.1 Irreducible Polynomials.................................. 26 8.2 Existence of Roots..................................... 26 9 Introduction to Finite Fields 27 1

1 RINGS 2 10 Formal Derivatives 30 11 Extension Fields 31 12 Iterated Field Extensions 31 13 Splitting Fields 32 14 Galois Group 34 15 More Field Extension Results 40 16 Discussion Questions about Fields 45 17 Galois Correspondence 47 18 Galois Theory for Cubic Polynomials 50 18.1 Solving the Cubic...................................... 50 19 Galois Theory of Quartic Polynomials 52 20 Solvability of Polynomials by Radicals 54 20.1 Insolvability of the Quintic................................ 56 21 Symmetric Functions 57 22 Other Descriptions of Galois Theory 57 Abstract These are informal notes taken from a variety of sources on basic ring theory and Galois theory. 1 Rings 1.1 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 3 Example 1.3. Matrices whose entries come from any of the previous examples are rings. Definition 1.2. If R 1, R 2,..., R n are all rings, then their direct product R 1 R n 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 Note: Z[i] is a subring of the complex numbers C. ring of Gaussian integers. 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 Z n is a zero divisors if and only if a is not relatively prime to n. Corollary 1.1. Z p 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, Z p, 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 4 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 Z n, we find U(Z n ) = 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. Z p is a field if and only if p is prime. Example 1.9. Q( 2) is a field. Example 1.10. Z 3 [i] and Z 7 [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 R be a homomorphism between two rings R and R. Then: 1. φ(0) = 0; 2. φ( a) = φ(a); 3. φ(na) = nφ(a);

2 RING HOMOMORPHISMS 5 4. φ is injective if and only if Ker(φ) = {0}; 5. φ(a) n = φ(a n ), for all n > 0; 6. Ker(φ) is a subring of R. Example 2.1. Consider the polynomial equation 2x 3 5x 2 + 7x 8 = 0. Claim: this equation has no integer solutions. We argue by contradiction. Let φ : Z Z 3 be the usual map x x mod 3. Suppose this equation does have an integral solution, say a. Then: 0 = φ(a) = 2a 3 5a 2 + 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 + b 2 + b + 1 = 0 for some b Z 3. By exhaustive checking, we find there is no such element b. Observation: Let φ be a homomorphism between two rings, say R 1 and R 2. Set a = φ(1). Then it is easy to check that a 2 = a. One can use this to show that the only ring homomorphisms of Z n 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 R 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 be a surjective homomorphism between two rings. Then R/I = R, 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.

2 RING HOMOMORPHISMS 6 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 x 0 [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 R and a A so 1 = br + a. Hence (b + I)(r + 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]/ < x 2 + 1 > which will be isomorphic to the field of complex numbers. Let g(x) R[x]. Then the coset g(x)+ < x 2 + 1 > can be represented as a 1 x + a 0 + < x 2 + 1 > by division of polynomials. Further x 2 = 1 in A (verify!). We can find the multiplicative inverse, say b 1 x+b 0 + < x 2 +1 >, of a non-zero element a 1 x+a 0 + < x 2 + 1 >. Set Then (a 1 x + a 0 + < x 2 + 1 >) (b 1 x + b 0 + < x 2 + 1 >) = 1+ < x 2 + 1 >. (4) a 0 b 0 a 1 b 1 = 1, a 1 b 0 + a 0 b 1. (5) To solve for a 0, a 1, we consider two cases: either b 0 0 or b 1 0. In both cases, we will find that a 0 = b 0 b 2 0 +, a 1 = b 1 b2 1 b 2 0 +. (6) b2 1 Further, we observe that the quotient ring is isomorphic to the field of complex numbers.

3 CHINESE REMAINDER THEOREM 7 Example 2.5. Consider A = Q[x]/ < x 2 2 >. Then in the quotient x 2 = 2. The cosets can be represented as b 1 x + b 0 + < x 2 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 2 2 + < 2 i >= i 2 + < 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 Z 5. 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 j a jb j 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 A 1, A 2,..., A k be ideals in R. Consider the mapping R R/A 1 R/A 2 R/A k by r (r + A 1, r + A 2,..., r + A k ) (7) is a ring homomorphism with kernel A 1 A 2... A k. If for each i, j {1, 2,..., k} with i j the ideals A i and A j are comaximal, then the map is surjective and A 1 A 2... A k = A 1 A 2... A k. Hence, we have the natural isomorphism R/(A 1 A 2... A k ) = R/(A 1 A 2... A k ) = R/A 1 R/A 2 R/A k. (8) Proof. We first show the case when k = 2. Consider the map φ : R R/A 1 R/A 2 defined by φ(r) = (r mod A 1, r mod A 2 ). (9)

3 CHINESE REMAINDER THEOREM 8 This map is a ring homomorphism since r r mod A 1 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 A 1 and r A 2 ; that is, r A 1 A 2. Note: all this holds without any restrictions on the ideals A 1 and A 2. To complete the proof, we now impose the condition A 1 and A 2 are also comaximal. We must establish that (1) φ is surjective, and (2) A 1 A 2 = A 1 A 2. The condition A 1 and A 2 are comaximal forces A 1 +A 2 = R. In particular, there must exist elements x A 1 and y A 2 such that x + y = 1. So φ(x) = (0, 1) and φ(y) = (1, 0) (verify). Now let r = (r 1 mod A 1, r 2 mod A 2 ) be an arbitrary element of the product R/A 1 R/A 2. We claim that the element r 2 x + r 1 y is mapped to r. Consider: φ(r 2 x + r 1 y) = φ(r 2 )φ(x) + φ(r 1 )φ(y) (10) = (r 2 mod A 1, r 2 mod A 2 ) (0, 1) + (r 1 mod A 1, r 1 mod A 2 ) (1, 0) (11) = (0, r 2 mod A 2 ) + (r 1 mod A 1, 0) (12) = (r 1 mod A 1, r 2 mod A 2 ). (13) Hence, the ring homomorphism φ is surjective. It remains to show that A 1 A 2 = A 1 A 2. Now, the ideal A 1 A 2 is always contained in the intersection A 1 A 2. If A 1 and A 2 are comaximal and x A 1 and y A 2 are chosen as above, then for any c A 1 A 2, we have c = c1 = cx + cy A 1 A 2. (14) Hence A 1 A 2 A 1 A 2. The general case follows by induction from the case of two ideals by using A = A 1 and B = A 2 A 3 A k once we know that A 1 and A 2 A 3 A k 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 Z m has ideals I =< p > and J =< q > with zero intersection. Hence Z m is isomorphic to the direct product of Z p and Z q. 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 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 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 REVIEW OF BASIC NUMBER THEORY 9 4 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: m 0 a + n 0 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 where p 1 > p 2 >... are prime and each α i > 0. a = p α1 1 pα2 2 pα k k, (15) 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. (Uniqueness) Consider a = p α1 1 pα2 2 pα k k = q β1 1 qβ2 2 qβ l k where p 1 > p 2 >... and q 1 > q 2 >... are prime and their exponents are only positive. Claim: k = l, p i = q i and α i = β i for all i. We use induction. The result holds for a = 2. We assume the result for all integers less than a. Since α 1 > 0, we find p 1 a so p 1 q β1 1 qβ2 2 qβ l k. (16)

4 REVIEW OF BASIC NUMBER THEORY 10 In particular, p 1 q i for some i, since p 1 is prime. But q 1 > q i = p 1. On the other hand, q 1 a implies q 1 p j. As before, p 1 p j q 1. Hence, q 1 p 1 and p 1 q 1 implies p 1 = q 1. Without loss of generality, assume α 1 b 1. We cancell out one factor of p 1. Then But induction, p i = q i, k = l and α i = b i for all i. a = p α1 1 1 p α2 2 p pα k k = p β1 1 1 q β2 2 qβ l k. (17) 1 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 a p 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)/2a 2 + + a p, 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 + a p )(modp). By induction, a p 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 a p 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). (18) 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 n 0 and m 0 so that 1 = n 0 b m 0 a. (21) We can simply multiply this last equation by α β to obtain the desired solution.

5 EUCLIDEAN DOMAINS 11 5 Euclidean Domains 5.1 Introduction Definition 5.1. An integral domain R is an Euclidean domain if for every a 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 {0}. Choose a 0 A, non-zero, so that d(a 0 ) is minimal. Let a A. By divison, we find a = qa 0 + r. By minimality, r = 0. Definition 5.2. Let a 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.

5 EUCLIDEAN DOMAINS 12 5.2 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 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 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 a 1 a 2 a n, then π will divides at least one of the elements a 1, 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.

5 EUCLIDEAN DOMAINS 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). (22) 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 (23) 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 = au 1. 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 0 be a non-unit in R. Suppose that a = π 1 π 2 π n = π 1π 2 π m where π i and π j are all prime elements in R. Then n = m and each π is an associate of some π j and conversely. Proof. Examine: a = π 1 π 2 π n = π 1π 2 π m. (24) We know that π 1 must divide one of primes π 1, π 2,..., π m, say π i. Hence, π 1 and π i are associates; that is, π i = u 1π 1. Next, we may cancel out the factor π 1 in the above products to obtain: π 2 π n = u 1 π 1π 2 π i 1π i+1π m. (25) We repeat this procedure n times to obtain the equation: 1 = u 1 u 2 u n z, where z is a certain product of the remaining primes π. We conclude that n m since the primes π are not units. We reverse the roles of the two factorizations to get m n. Hence, m = n and each prime π i is an associate of π j.

5 EUCLIDEAN DOMAINS 14 Proposition 5.10. Let ideal A =< a 0 > is maximal in a Euclidean domain R if and only if a 0 is a prime element. Proof. We first show that if a 0 is not prime, then A =< a 0 > is not a maximal ideal. Write a 0 = bc where b, c R and neither b nor c are units. Let B =< b >. Then a 0 B so A B. To finish we need to verify that A B and B 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 = xa 0 for some x R. But a 0 = bc so b = bcx 0. By cancellation, 1 = cx 0. So c must be a unit. Again, this contradicts our assumption on c. Next, we assume that a 0 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 =< u 0 >. Since a 0 A U =< u 0 >, we find a 0 = xu 0 for some x R. Since a 0 is prime, we know that either x or u 0 must be a unit and the remaining element is equal to a 0. If u 0 is the unit, then U = R. If x is the unit, then u 0 = x 1 a 0 A. Hence, U = A. 5.3 Gaussian Integers For Z[i], we define d(z) = d(a + bi) = a 2 + b 2. Proposition 5.11. Z[i] is an euclidean ring. Proof. We just have to verify the divison algorithm: y = tx + r (26) 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 + u 1 and b = vn + v 1 where u 1 and v 1 are integers satisfying u 1 1 2 n and v 1 1 2 n. Let t = u + vi and r = u 1 + v 1 i. Then y = a + bi (27) = un + u 1 + (vn + v 1 )i (28) = (u + vi)n + (u 1 + v 1 i) (29) = tn + r (30) Since d(r) = d(u 1 + v 1 i) = u 2 1 + v 2 1 n 2 /4 + n 2 /4 < n 2 = d(n), the result holds for this special case. For the general case, we assume x 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). (32)

5 EUCLIDEAN DOMAINS 15 Since x 0 and d(x) is a positive integer, we may deduce: d(y tx) < d(x). (33) We may write y = tx + r 0 where r 0 = y tx. So t and r 0 are Gaussian integers. Then either r 0 is 0 or d(r 0 ) = 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 = w 1 w 2 in Z[i] then p = N(w 1 )N(w 2 ) so one of N(w 1 ) or N(w 2 ) 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 so p = ππ in Z[i]. In other words, the prime integer p factors in the larger ring Z[i]. Now N(π)N(π ) = N(p) = p 2. Since π is not a unit, there are two possibilities for the norms: either N(π) = ±p 2 or N(π) = ±p. In the former case N(π ) = ±1 hence π must be a unit and p = π (up to associates) is irreducible in Z[i]. In the latter case, N(π) = N(π ) = ±p. Hence π is also irreducible and p = ππ is a product of two irreducibles. In particular, if we write π = a + bi so N(π) = a 2 + b 2, we find p = N(π) = a 2 + b 2. 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 = a 2 + b 2 is the sum of two integer squares. If p = a 2 + b 2, 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 = a 2 + b 2. If both a 2 and b 2 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 1 2 + 1 2 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 1 2 + 1. If p is an odd prime, then p divides n 2 + 1 is equivalent to n 2 = 1 in Z p. 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 Z p since ( 1) 2 = 1. Hence p divides an integer of the form n 2 + 1 if and only if the multiplicative group Z p has an element of order 4.

5 EUCLIDEAN DOMAINS 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 n 2 + 1. To see this, consider n 4 1 mod p in the group Z p. Then p must divide n 4 1 or the product (n 2 1)(n 2 + 1) = (n + 1)(n 1)(n 2 + 1). Since n < p 1, we find p must divide n 2 + 1. Now to the proof that Z p contains an element of order 4. We first show that Z p contains a unique element of order 2. Consider m 2 1 mod p so m 2 1 0 mod p. Then p will divide m 2 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 Z 2 Z 2 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 n 2 + 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 = a 2 + b 2 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 p 2, (c) a + bi, a bi, the distinct irreducible factors of p = a 2 + b 2 = (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 x 2 + 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 uα 2 where u is a Gaussian unit and α is a Gaussian integer. Write α itself as m + ni, then we may write: {x, y} = {±(m 2 n 2 ), ±2mn}, z = ±(m 2 + n 2 ). (34) 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.

5 EUCLIDEAN DOMAINS 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 N : R Z, a + b 2 a 2 2b 2. (35) Then it is straightforward to verify that N(z 1 z 2 ) = N(z 1 )N(z 2 ), for all z 1, z 2 R. For the division algorithm, introduce d(z) as N(z). Let x = a + b 2 and y = c + d 2 where y 0. Then x y = 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 ( ) x x = qy + y y q. (36) Then the remainder is z = x qy or z = y ( x y ). q We find that N(z) = ( ) x N(y)N(y y q (37) = N(y) ( (α r) 2 2b 2) (38) Now ( (α r) 2 2b 2) 2/4 = 1/2 so d(z) (1/2)d(y) < d(y) as is required. Identical reasoning with 3 gives ( (α r) 2 3b 2) 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 a 2 2b 2 = ±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 {±u k : 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 cube root of unity so ω = 1 2 + 3 2 i. We can introduce a norm N by: N : Z[ω] Z, a + bω a 2 ab + b 2. (39)

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) = u 2 + v 2. Z[ω] has only six unit elements. We can introduce a division algorithm by imitating yet again the Gaussian integers. 6 Example of PID that is not a Euclidean Domain 6.1 How to show something is not a Euclidean Domain Let D be an integral domain with unit and let D = {0} {unitsin D}. Then D is a field if and only if D = D. Proposition 6.1. Let D be a Euclidean domain with absolute value function φ : D Z +. Assume that D is not a field, and choose w D \ D so that φ(w) is minimal among φ(d \ D). Then 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 x = qw + z, with q, x D and φ(z) < φ(w). By hypothesis on φ(w), z D, and qw = x z. Corollary 6.1. If R = {a + b(1 + 19)/2 : a, b Z}, then R is not a Euclidean domain. Proof. It is enought to show that there is no w R \ R such that for all x R, w divides x z for 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 = a 2 + ab + b 2 /4 + 19b 2 /4 = a 2 + ab + 5b 2 Z, so that if z is a unit, a 2 + ab + 5b 2 = 1 and so b = 0 and a = ±1. Thus R = {0, 1, 1}. 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 {0} be an ideal of R. Choose y A with y > 0 minimal (possible since y 2 = 1, 2,... if y 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.

7 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 ( ) x z w = a + b 19 y c (d + e 19 ) (q f 19) = r + 19. c This is non-zero with absolute value r2 +19 c 1 since r c/2 and c 5. 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 x y z w = 1 2 < 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 a 2 + 19b 2 a 2 + b 2 mod 3. Let z = a b 19 and choose w Z so that a 2 + 19b 2 = 3w + r with r = 1 or 2. Then x y z w = a 2 +19b 2 3 w = r 3 < 1. Case 4: c = 4. Since our fraction is in lowest terms, a and b are not both even. If they are both odd, a 2 + 19b 2 a 2 + 3b 2 1 + 3 4 mod 8, so we can choose w Z with a 2 + 19b 2 = 8w + 4. Let z = a b 19 2 R. Then x y z w = a 2 +19b 2 8 w = 1 2 < 1. If they are of opposite parity, a 2 + 19b 2 a 2 b 2 mod 4, so we can choose w Z with a 2 + 19b 2 = 4w + r, with r = 1, 2, or 3. Let z = a b 19. Then x y z w = a 2 +19b 2 4 w = r 4 < 1. 7 Ring of Polynomials 7.1 Introduction Definition 7.1. Let R be a ring. A polynomial with coefficients in R and indeterminate x is a finite sum n f(x) = a i x i = a n x n + a n 1 x n 1 + a 1 x + a 0 (40) where a i R. i=0 We add polynomials by adding the coefficients of similar powers. Multiplication is given as follows. Let p(x) = a 0 +a 1 x+ +a m x m and q(x) = b 0 +b 1 x+ +b n x n, then p(x)q(x) = c 0 +c 1 x+ +c k x k where c t = a t b 0 + a t 1 b 1 + + a 0 b t. We find that F [x] is a commutative ring with unity. If p(x) = a 0 + a 1 x + + a m x m 0 and a m 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) = a 0 + a 1 x + + a m x m and q(x) = b 0 + b 1 x + + b n x n where a m 0 and b n 0. Write p(x)q(x) = c 0 + c 1 x + + c k x k where c t = a t b 0 + a t 1 b 1 + + a 0 b t. We claim that c t = 0 for t > m + n while c m+n 0. It is easy to check that c m+n = a m b n 0 since F is a field. Next we consider c t where t > m+n. Examine the term in its definition a i b j where i+j = t > m+n or j = t i. If i > m + n then a i = 0 so a i b j = 0. Suppose that i m + n. Then j = t i

7 RING OF POLYNOMIALS 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 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) 0. We need to establish the divison algorithm. Proposition 7.3. Given two polynomials p(x) and q(x) where q 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) = a 0 + a 1 x + + a m x m (41) q(x) = b 0 + b 1 x + + b n x n (42) where a m 0 and b n 0. Let p 1 (x) = p(x) (a m /b n )x m n q(x); then deg p 1 < deg p. By induction on the degree of p we find p 1 (x) = t 1 (x)q(x) + r(x) where either r(x) = 0 or deg(r) < deg(q). In other words, p(x) = p 1 (x) + (a m /b n )x m n q(x) = t 1 (x)q(x) + (a m /b n )x m 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.

7 RING OF POLYNOMIALS 21 Proof. We use induction on the degree of f. We use the factor theorem. If f has degree 1, then f(x) = a 0 + a 1 x. But f(x) = a 0 (a 1 x/a 0 + 1). Then f(x) is zero if and only if a 1 x/a 0 + 1 = 0. We find x = a 0 /a 1. 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) = f 1 (x)(x α). By induction, f 1 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) = c 0 + c 1 x where c 1 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) = a 0 + a 1 x + + a n x n (43) 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 a 0 and s divides a n in Z. Proof. We begin by writing: By multiplying by s n, we obtain: 0 = f(a) = a n (r/s) n + a n 1 (r/s) n 1 + + a 1 (r/s) + a 0. (44) a n r n + a n 1 r n 1 s + + a 1 rs n 1 + a 0 s n = 0. (45) By solving for a n r n and, separately, a 0 s n, we obtain two equations: a n r n = s[a n 1 r n 1 + + a 1 rs n 2 + a 0 s n 1 ] (46) a 0 s n = r[a n r n 1 + + a 1 s n 1 ]. (47) Since r and s are relatively prime, we find s divides a n and r will divide a 0.

7 RING OF POLYNOMIALS 22 Definition 7.3. Let f(x) = a n x n + a n 1 x n 1 + + a 1 x + a 0 be a polynomial in Z[x]. Then c = gcd(a n,..., a 0 ) 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]. f(x)g(x) is also primitive. Then their product 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] Z p [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 Z p [x]. However, Z p [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 Z p [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 where c(f) Q is positive and f (x) Z[x] is primitive. f(x) = c(f)f (x) (48) Proof. We write f(x) as a 0 /b 0 + (a 1 /b 1 )x + + (a n /b n )x n Q[x]. Let B = b 0 b 1 b n, so f(x) = (1/B)g(x) where g(x) Z[x]. Next, define B as the content of g(x) which is positive. Then f(x) = c(f)f (x) where c(f) = B /B and f (x) = (B/B )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 Proof. We have: c(f) = c(g)c(h) and f (x) = g (x)h (x). (49) f(x) = g(x)h(x) (50) = [c(g)g (x)] [c(h)h (x)] (51) = c(g)c(h) g (x)h (x). (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].

7 RING OF POLYNOMIALS 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 Z p. 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[x 1, x 2,..., x n ]. 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 p n = 1 = a r b s ; that is, a r and b s are units in D. Hence p(x) = [b s a(x)] [a r b(x)] where b s a(x) and a r 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) = x 2 + 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 a 1, a 2, a 3,... such that each a i is divisible by a i+1 and a i and a i+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 < a i > which are nested < a 1 > < a 2 >. (53) We notice that their union i=1 < a i > 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 < a i >, say i = i 0. Then < a i >=< a i0 > for all i i 0. 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)).)