# Rings. Chapter Homomorphisms and ideals

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1 Chapter 2 Rings This chapter should be at least in part a review of stuff you ve seen before. Roughly it is covered in Rotman chapter 3 and sections 6.1 and 6.2. You should *know* well all the material in chapter 3 of Rotman already. 2.1 Homomorphisms and ideals We recall that a ring R is a set possessing two operations + and. such that it is an Abelian group under + and a monoid under. satisfying the additional axioms: a(b+c) = ab+ac, (b+c)a = ba+ca for all a, b, c R. Also, a homomorphism f : R S is a morphism of Abelian groups such that f(ab) = f(a)f(b) for all a, b R and f(1 R ) = 1 S. I warn you: all rings and all ring homomorphisms are unital throughout these notes. This is the right convention for algebra but the wrong convention for analysis... This is an issue you need to be flexible on! There are many minor variations on the definitions in ring theory which you just have to live with. A ring is called commutative if in addition ab = ba for all a, b R. A division ring is a ring R for which R is a group under multiplication, i.e. 1 0 and every non-zero element of R is a unit. A field is a commutative division ring. Let me give you the standard example of a division ring that is not a field: the quaternions. Start with the real vector space on basis {1, i, j, k}. Define multiplication by (a 1 + a 2 i + a 3 j + a 4 k)(b 1 + b 2 i + b 3 j + b 4 k) = (a 1 b 1 a 2 b 2 a 3 b 3 a 4 b 4 ) + (a 1 b 2 a 2 b 1 + a 3 b 4 a 4 b 3 )i + (a 1 b 3 + a 3 b 1 + a 4 b 2 a 2 b 4 )j + (a 1 b 4 + a 4 b 1 + a 2 b 3 a 3 b 2 )k. In other words, the operation is R-bilinear, and i 2 = j 2 = k 2 = 1, ij = k, jk = i, ki = j, ji = k, kj = i.ik = j. Note the elements {±1, ±i, ±j, ±k} form a subgroup of the group of units of the quaternions, and this group of order 8 is the quaternion group Q 3. A ring is called an integral domain if it is a commutative ring with 1 0 such that there are no non-zero zero divisors. Here, a zero divisor in a commutative ring is an element z R for which zy = 0 for some y R. In other words, a commutative ring with 1 0 is an integral domain if ab = 0 implies either a = 0 or b = 0. For example, Z is integral domain, while Z n is an integral domain if and only if n is prime. Let R be a ring. A basic notion is the characteristic of R. This is defined to be the smallest positive integer n such that n.1 R = 1 R R (n times) = 0, or else 0 if no such positive integer exists. For instance, Z is a ring of characteristic 0, while Z n for n > 1 has characteristic n. The zero ring has characteristic 1. Note that there is a unique ring homomorphism Z R, n n.1 R. The integer char R generates the kernel of this homomorphism. A subring of a ring R is a subset S of R that is closed under addition and multiplication and which is a ring in its own right with the induced operations. 41

2 42 CHAPTER 2. RINGS Note we do not insist that a subring of R contains 1 R : it has to be unital but the unit 1 S in the subring S may be different from the unit in the ring R. Beware, though, that the natural inclusion map i : S R is not (in these notes) a ring homomorphism if 1 S 1 R! In case S contains 1 R (i.e. 1 R = 1 S ) we will call S a unital subring, in which case the inclusion map is a ring homomorphism. In general, an element e of R is called an idempotent if it has this property e 2 = e. In that case, ere = {eae a R} is a subring of R with identity e. If S is a subring of R, then 1 S satisfies 1 2 S = 1 S, i.e. 1 S is an idempotent, and then S is a unital subring of the not necessarily unital subring 1 S R1 S of R. This gives us plenty of examples of subrings: (1) Take T = R S for any two rings R, S (the product in the category of rings: the Cartesian product as a set with coordinatewise addition and multiplication). Let e R = (1 R, 0) T and e S = (0, 1 S ) T. Then, e R and e S are idempotents and R = e R T e R, S = e S T e S. (2) Let R be any ring and M n (R) be the ring of n n matrices with entries in R, under ordinary addition and multiplication of matrices. Then, e i, the matrix with a 1 R in the ith diagonal entry and zeros in all other entries, is an idempotent in M n (R) and R = e i M n (R)e i. If f : R S is a homomorphism, its image is a unital subring of S. The kernel is not (in general) a subring of R (because it may not contain a 1), but it is something else: an ideal. By definition, an ideal is a sub-abelian group I of R such that for all a A, x R, we have that both ax and xa are elements of I. Sometimes, we will call an ideal a two-sided ideal for emphasis. There are also a weaker notion of a left (resp. a right) ideal, namely, a sub-abelian group I of R such that ax I (resp. xa I) for every a A, x I. We will discuss left and right ideals in more detail when we come to study modules. In a commutative ring, left, right and two-sided ideals are all the same thing. Let me now introduce a convenient notation. Let X, Y R. Instead of writing XY for the set {xy x X, y Y } as in group theory, I will write XY for the sub-abelian group of R generated by the elements {xy x X, y Y }. So elements of XY look like linear combinations ±x 1 y 1 ± ± x n y n for n 0 and x i X, y i Y. Then, the statement that I is an ideal (meaning a two-sided ideal always) of R can be stated simply as saying that RIR = I. In general, given a subset X R, the ideal generated by X is the set RXR (check that it is an ideal!). We will denote RXR instead by (X). It is the unique smallest ideal of R containing the set X. You need to be careful: these statements all depend on R being unital, and get more complicated for non-unital rings. Now, the set of ideals of R is a complete lattice. Indeed, given a family of ideals I j (j J), their greatest lower bound is simply their intersection, also an ideal. Their least upper bound is the ideal generated by the given ideals, which is simply the set of all elements of R of the form j J i j for elements i j I j all but finitely many of which are zero. In other words, the ideal generated by ideals I j (j J) is j J I j. There is one other way of building new ideals out of old: products. Given two ideals I, J of R, note that IJ is also an ideal, clearly contained in both I and J hence contained in I J. We have constructed ideals IJ I J I, J I + J. Finally, let me note that if I is an ideal of R then I = R if and only if I contains a unit. Let R be a ring and I be an ideal. Let R/I be the quotient Abelian group, so elements of R/I are cosets {a + I a R}. We make R/I into a ring in its own right by defining the multiplication by (a + I)(b + I) = ab + I. This is well-defined precisely because I is an ideal. The unit of the quotient ring R/I is the element 1 R + I. The map π : R R/I, a a + I is a homomorphism, called the quotient map. Univeral property of quotients. Let I be an ideal of R and f : R R be a homomorphism such that ker f I. Then, there exists a unique homomorphism f : R/I R such that f = f π. Corollary. Let f : R R be an epimorphism with kernel I. Then, f factors through the quotient R/I to induce an isomorphism f : R/I R.

3 2.2. MAXIMAL AND PRIME IDEALS 43 We also record: Lattice isomorphism theorem. Let f : R R be an epimorphism with kernel I. Then, the map J f(j) gives an isomorphism between the lattice of ideals of R containing I and the lattice of ideals of R. 2.2 Maximal and prime ideals We call an ideal I of a ring R proper if I R. Note the zero ring has no proper ideals! An ideal M of a ring R is called a maximal ideal if it is maximal amongst all proper ideals of R. Thus, if M is a maximal ideal, then M R and whenever M J R for an ideal J, then either J = M or J = R. An ideal P of a ring R is called a prime ideal if P R and for all ideals A, B of R, the property AB P implies either A P or B P. Characterization of maximal ideals. Let M be an ideal of R. (i) If M is maximal and R is commutative, then R/M is a field. (ii) If R/M is a division ring, then M is maximal. Proof. (i) (ii). If M is maximal and R is commutative, the lattice isomorphism theorem implies that R/M is a commutative ring with 1 0 and no non-zero proper ideals. Thus, if we take any element 0 x R/M, we must have that (x) = R/M, hence x is a unit. (ii) (i). If R/M is a division ring, then R M and R/M has no non-zero proper ideals. Hence, the zero ideal in R/M is a maximal ideal. Now the lattice isomorphism theorem implies that M is a maximal ideal of R. Thus: for commutative rings, M is maximal if and only if R/M is a field. Characterization of prime ideals. Let P be a proper ideal of R. Then: (i) If ab P implies either a P or b P for all a, b R, then P is a prime ideal. (ii) If R is commutative and P is prime, then ab P implies either a P or b P for all a, b R. Proof. (i) (ii). Take ideals A, B of R with AB P. If A P, we are done. So assume A P and take a A P. For every b B, we have that ab AB P, hence by hypothesis b P. This shows B P as required. (ii) (i). Let P be prime and take elements a, b R with ab P. Then, (ab) P. In a commutative ring, (ab) = (a)(b) (proof: (a)(b) = RaRb = Rab = (ab) as R is commutative). So, (a)(b) P. So one of (a) or (b) lies in P, hence one of a or b is in P. Thus: for commutative rings, P is prime if and only if R/P is an integral domain. A special case of this gives that a commutative ring is an integral domain if and only if (0) is a prime ideal. Corollary. In a commutative ring, all maximal ideals are prime ideals. Proof. We apply the characterizations. Suppose M is maximal. Then, R/M is a field, hence an integral domain, hence M is prime. Consider R = Z. Then, ideals are in the first place Abelian subgroups, so equal (n) (all multiples of n) for some n 0. Each (n) is indeed an ideal. Hence, the ideals of Z are the (n) for n 0. An ideal (n) is prime if and only if n is a prime number or 0; an ideal (n) is maximal if and only if n is a prime number. Observe: {prime ideals} = {maximal ideals} {(0)}. We turn next to proving an important theoretical result. The proof depends crucially on Zorn s lemma. Since this is the first time we have used this, let me give a review first. Let (A, ) be a non-empty partially ordered set. Call a chain in A a non-empty subset B that is totally ordered by. (Warning: B may be uncountable!!) Then:

4 44 CHAPTER 2. RINGS Zorn s lemma. If every chain B in A has an upper bound in A, i.e. there exists an a A such that b a for all b B, then A contains a maximal element. Now we apply it to prove existence of maximal ideals: Theorem. Let R be a non-zero ring. Then, every proper ideal of R is contained in a maximal ideal. In particular, maximal ideals exist. Proof. Let A = {proper ideals J of R with I J}, partially ordered by. Take a chain B = (K ω ) ω Ω of ideals in A. Then, K = ω Ω K ω is an ideal of R. Moreover, if 1 K then 1 K ω for some ω Ω, contradicting the assumption that all K ω are proper ideals of R. Hence, K is a proper ideal of R, and clearly K contains I, so K A. This verifies that every chain in A has an upper bound in A. Now we apply Zorn s lemma to deduce that A contains a maximal element M. Then, M is exactly a maximal ideal of R containing I. Warning: It is absolutely crucial here that we are working with unital rings. There exist non-unital rings with no maximal ideals!!! The last topic in this section is the Chinese remainder theorem. Recall that if R i (i I) are rings, their product (in the category of rings) is the Cartesian product i I R i with coordinatewise addition and multiplication. The product comes with the natural projections π i : i I R i R i which are ring homomorphisms. Chinese remainder theorem. Let I 1,..., I n be ideals of R such that I j + I k = R for all j k. Let π j : R R/I j be the canonical quotient map and define π : R R/I 1 R/I 2 R/I n, r (π 1 (r), π 2 (r),..., π n (r)). Then, π has kernel I 1 I n, and the induced map is an isomorphism. π : R/I 1 I n R/I 1 R/I n Proof. We first show by induction on m = 2,..., n that R = I 1 + (I 2 I m ). The case m = 2 is immediate from the hypothesis of the theorem. For m > 2, induction gives that R = I 1 + (I 2 I m 1 ). Hence, R = R 2 = (I 1 + (I 2 I m 1 ))(I 1 + I m ) I 1 + (I 2 I m 1 )I m I 1 + (I 2 I m 1 I m ) R This verifies the induction hypothesis. Now we prove the theorem. It is easy to check that the kernel of π is I 1 I n. So we just need to prove that π is surjective. In other words, we need to show that given r 1,..., r n R, there exists r R with r + I j = r j + I j for all j = 1,..., n. For each j, write I j = k j I k. By the opening paragraph, we know R = I j + I j. So we can find an element s j I j such that r j + I j = s j + I j. Set r = s s n. Then, r + I j = s j + I j = r j + I j for each j, as required. For example, consider the case R = Z. Let n 1,..., n a be integers with (n i, n j ) = 1 for i j ( pairwise coprime ). Then, Z = (n i ) + (n j ) for i j, verifying the hypothesis in the theorem. So, the theorem gives a ring isomorphism Z n1...n a = Zn1 Z na. Thus, the Chinese remainder theorem generalizes Lemma

5 2.3. EUCLIDEAN DOMAINS AND PIDS Euclidean domains and PIDs In this section, all rings will be commutative. A principal ideal domain (PID) is an integral domain in which every ideal is principal, i.e. look like (a) = Ra for some a R. For example, we have already seen that in Z, all ideals have the form (n) for n 0, so Z is a PID Lemma. If R is a PID, then all non-zero prime ideals are maximal ideals. Proof. Let (a) be a non-zero prime ideal, and suppose (a) (b). Then a = bc for some c R, hence, (b + (a))(c + (a)) = 0 in the integral domain R/(a). This means that either b + (a) = 0, whence b (a) whence (a) = (b), or c + (a) = 0. In this latter case, c (a) so c = ad for some d R. But then, a = bc = bad so bd = 1 and b is a unit, i.e. (b) = R. This shows (a) is a maximal ideal. A Euclidean domain is an integral domain R equipped with a degree function φ : R N such that (E1) If a, b R then φ(a) φ(ab); (E2) If a R, b R then there exists q, r R such that with either r = 0 or r 0 and φ(r) < φ(b). a = qb + r For example, Z is a Euclidean domain, setting φ(n) = n (absolute value). Condition (E2) for Z is exactly the division algorithm. ED PID. If R is a Euclidean domain, then R is a PID. Proof. Take a non-zero ideal I of R. Pick 0 a I with φ(a) minimal. We claim simply that I = (a). Indeed, take any other 0 b I. Write b = qa + r with r = 0 or φ(r) < φ(a). Then, r = b qa I. The minimality of the choice of a implies that in fact r = 0. Hence, b = qa so b (a). Now we can use the notion of an Euclidean domain to give more examples of PIDs. (By the way, any field is a Euclidean domain in a silly way; think degree function φ works!) The first example is the polynomial ring in an indeterminate X over a field. Let s first define polynomial rings more generally. So suppose R is any commutative ring. Then, the ring R[X] (X an indeterminate) consists of all sequences (a 0, a 1,... ) with each a i R and a n = 0 for all n >> 0. The addition in R[X] is just the coordinatewise addition of the sequences. But multiplication is defined by convolution : where (a 0, a 1,... )(b 0, b 1,... ) = (c 0, c 1,... ) c n = i+j=n for each n 0. There is a little checking needed here to verify this is indeed a ring! We henceforth write a 0 + a 1 X + a 2 X as shorthand for the element (a 0, a 1,... ), bearing in mind that all but finitely many of the a i are zero so it makes sense to write down such an infinite sum. Now, given a polynomial f = a 0 + a 1 X + R[X], we define its degree deg f to be the largest n 0 such that a n 0. It is also convenient to define the degree of the zero polynomial to be. a i b j

6 46 CHAPTER 2. RINGS Then, we have that deg(f + g) max(deg f, deg g) and deg(fg) deg f + deg g for all f, g R[X]. In fact, providing R is an integral domain, deg(fg) = deg f + deg g, which shows that R[X] is also an integral domain in that case. Now we have the familiar division algorithm for polynomials: Division algorithm for polynomials. Let R be a commutative ring, f, g R[X]. If the leading coefficient of g is a unit, then there exist unique polynomials q, r R[X] such that with deg r < deg g. f = qg + r Proof. Let f = a n X n + + a 1 X + a 0 and g = b m X m + + b 1 X + b 0 with a n 0 and b m a unit. We first prove existence of r and q by induction on n, the conclusion being trivial if n m. If n > m, then deg(f a n b 1 m X n m g) < deg f = n. So by induction, there exist q, r R[X] with f a n b 1 m X n m g = q g + r with deg r < deg g. Thus, q = q + a n b 1 m X n m does the job. Now for uniqueness, suppose q, q, r, r R[X] with deg r deg r < deg g and qg +r = q g +r. Then, (q q )g = r r. So since b m is a unit, hence deg(q q ) = hence q = q, r = r. This immediately implies: deg(q q ) + deg g = deg(r r) < deg g Corollary. If F is a field, then F [X] is a Euclidean domain with degree function deg. Hence, F [X] is a PID. Let me give one other example of a Euclidean domain. Let ω C and define a ring homomorphism ev : Z[X] C, f(x) f(ω) evaluation at X = ω. Then, Z[ω] is defined to be the image of ev, a unital subring of C. So, Z[ω] = {f(ω) C f(x) Z[X]}. In general, Z[ω] is not a PID, but sometimes you get lucky... For instance, take ω = i, a square root of 1 in C. Then, Z[i] the ring of Gaussian integers is a Euclidean domain. The degree function φ = N is defined by N(a + ib) = a 2 + b 2. It is a good exercise to prove that this degree function really does make Z[i] into a Euclidean domain. Slightly more generally, consider the ring Z[ n] for an integer n. There is a function N : Z[ n] Z more usually called the norm defined by N(a + b n) = a 2 nb 2. In case n = 1, this is exactly the degree function making Gaussian integers into a Euclidean domain. In general, N may or may not be a degree function making Z[ n] into a Euclidean domain. But it is always the case that N is multiplicative, i.e. N(xy) = N(x)N(y) for all x, y Z[ n]. This is very useful, for instance:

7 2.3. EUCLIDEAN DOMAINS AND PIDS Lemma. x Z[ n] is a unit if and only if N(x) = ±1. Proof. If x is a unit, then xy = 1 for some y so N(xy) = N(x)N(y) = N(±1) = 1. So N(x) is a unit in Z, hence equals ±1. Conversely, if N(x) = ±1, then x = a + b n with a 2 nb 2 = ±1. Then, ±(a + b n) a b n a 2 nb 2 = 1 so ± a b n a 2 nb 2 Z[ n] is the inverse of a + b n. For example, you can easily show now that the units in the ring Z[i] are exactly ±1, ±i. Let R be a commutative ring. We say an element a R divides b R, written a b, if there exists x R such that ax = b. Observe 1 divides everything and everything divides zero (although not everything is a zero divisor which we defined earlier, which is confusing). Define an equivalence relation associates on R by a ass b if a b and b a. Here are its basic properties: (1) a b if and only if (b) (a); (2) a ass b if and only if (a) = (b); (3) u R is a unit if and only if u r for all r R; (4) if a = bu for u a unit, then a ass b; (5) if R is an integral domain, then a ass b if and only if a = bu for u a unit. We call an element c R irreducible if c is non-zero and not a unit, and whenever c = ab for a, b R, then either a or b is a unit. We call an element p R a prime if p is non-zero and not a unit, and whenever p ab for a, b R then p a or p b Lemma. (i) p is prime if and only if (p) is a non-zero prime ideal. (ii) If R is an integral domain and (c) is a non-zero maximal ideal, then c is irreducible. (iii) If R is a PID and c is irreducible, then (c) is a non-zero maximal ideal. (iv) In PIDs, primes and irreducibles are the same thing. Proof. (i) This is immediate from the characterization of prime ideals. (ii) Suppose (c) is a non-zero maximal ideal and c = ab with a not a unit. Then, (c) (a) R, whence (c) = (a) as (c) is maximal. Thus, a = cd for some d R and we get c = ab = cbd. Hence, as we are in an integral domain, 1 = bd so b is a unit. (iii) Suppose c is irreducible. Then, (c) is a non-zero ideal. Let (c) (a) R for some a. Then, c = ab for some b. Hence, either a is a unit, in which case (a) = R, or b is a unit, in which case (a) = (c). Hence, (c) is a maximal ideal. (iv) If c is irreducible, then by (iii), (c) is a non-zero maximal ideal, hence a non-zero prime ideal and c is prime by (i). Conversely, let c be prime. Then, (c) is a non-zero prime ideal by (i), so by Lemma 2.3.1, (c) is a non-zero maximal ideal. Now (ii) gives that c is irreducible. Let me now give an important example of an integral domain that is not a PID. Consider R = Z[ 10]. We know by Lemma that a + b 10 is a unit if and only if a 2 10b 2 = ±1. I first show that 2 is irreducible in R. Indeed, if 2 = (a + b 10)(c + d 10) for integers a, b, c, d, then N(2) = 4 = (a 2 10b 2 )(c 2 10d 2 ). Assuming neither of a + b 10 or c + d 10 is a unit, this gives that a 2 10b 2 = ±2. Hence we get that a 2 2 or 3 (mod 5). But neither 2 or 3 is a quadratic residue modulo 5, so this is a contradiction. Now I show that 2 is not a prime in R. Indeed, 2 6 = (4 + 10)(4 10). But 2 (4 + 10) and 2 (4 10). Hence, 2 is an irreducible that is not prime, so Z[ 10] cannot be a PID by Lemma 2.3.3(iv).

8 48 CHAPTER 2. RINGS 2.4 Unique factorization domains Continue with all rings being commutative. A commutative ring R is a unique factorization domain (UFD) if (U1) R is an integral domain; (U2) every non-zero non-unit a R can be written as a = c 1 c 2... c n with each c i irreducible; (U3) If c 1 c 2... c n = d 1 d 2... d m for irreducibles c i, d i then n = m and (after reordering) c i ass d i for each i. Before proving the main result, we record an easy lemma (in language not defined yet, it says PIDs are noetherian ) Lemma. Let R be a PID and I 1 I 2 I 3... be an ascending chain of ideals in R. Then, for some n, I N = I n for all N n. (We say the ascending chain stabilizes.) Proof. Let I = j 1 I j, also an ideal of R. Then I = (a) for some a R as R is a PID. But then for some n 1 we actually have that a I n, hence I = I n, hence I N = I n for all N n. Now for the important theorem: PID UFD. If R is a PID then R is a UFD. Proof. Let R be a PID. Uniqueness. Suppose c 1... c n = d 1... d m for c i, d i R irreducible and n m. Then, c 1 is prime by Lemma 2.3.3(iv). So as c 1 d 1... d m, c 1 divides some d i, without loss of generality, c 1 d 1. So, d 1 = c 1 u where u is a unit, since d 1 and c 1 both irreducible. Now cancel c 1 and d 1 to get c 2... c n = d 2... d m and repeat. Let me clarify the last step. Then, you have c n = d n... d m and the argument gives c n = d n u for some unit, completing the proof in case m = n. But if m > n, then d n+1... d m is a unit. This means R = (d n1... d m ) (d n+1 ) so d n+1 is a unit too. But d n+1 is irreducible. Existence. The difficult part is proving existence of factorization. So suppose for a contradiction that R is a PID in which not every non-zero non-unit can be factorized into irreducibles. Call a R bad if it is a non-zero non-unit which cannot be factorized into irreducibles, and good otherwise. So by assumption, we can find a bad a R. Then, a is not itself irreducible. So as a is bad we can write a = a 1 b 1 with neither a 1, b 1 being units. If both a 1 and b 1 are good, we can factorize each of them both hence a. So without loss of generality, a 1 is bad. Now repeat, writing a 1 = a 2 b 2 with a 2 bad and b 2 not a unit, a 2 = a 3 b 3 with a 3 bad and b 3 not a unit, and so on. By the assumption that a is bad, this process must go on forever. Now set I j = (a j ). Then we have a chain I 1 I 2 I 3... with I j I j+1 for all j since (b j+1 ) is not a unit. This contradicts Lemma Remarks. (1) Strictly speaking, this argument is not quite logically correct. We secretly applied the Axiom of Choice (or Zorn s lemma). Hungerford gives the harder but logically correct proof. (2) It is hopefully obvious to you that the existence half of the proof of the theorem can be given for integral domains satisfying something weaker than being a PID: all that was needed was that the ring satisfied the ascending chain condition of Lemma 2.4.1, i.e. that the ring was noetherian. (3) In an integral domain in which factorizations exist, the uniqueness of factorization is actually equivalent to the statement that primes and irreducibles coincide. So in particular, in any UFD, primes and irreducibles are the same, as in PIDs. Once we know that R is a UFD (for instance if it is an ED or a PID), we can introduce the familiar notion of greatest common divisor. Actually, in any commutative ring R, an element d R is called a greatest common divisor of a, b R if d a, d b and moreover, whenever c a, c b then c d. It is obvious that if a GCD exists for elements a, b, then it is unique up to associates. Then:

9 2.5. LOCALIZATION OF RINGS 49 (1) If R is a UFD, then every pair of elements a, b R possess a greatest common divisor. I just sketch the idea here. If either a or b is zero, the GCD is 0; if either a or b is a unit, the GCD is 1. Otherwise, factorize a and b into irreducibles. Then, the GCD is c nc where the product is over all non-associate irreducibles c in R, where n c is the largest integer such that c nc a and c nc b. (2) If in fact R is a PID, then it is true that the GCD of a and b can be written in the form ra + sb for r, s R. This need not be the case in a general UFD. (3) Finally if in fact R is an ED, then the Euclidean algorithm gives an explicit procedure given a, b for computing their greatest common divisor in the form ra + sb. You should be familiar with applying this algorithm. 2.5 Localization of rings Let R be a commutative ring. A subset S of R is called multiplicative if 1 R S and S is closed under multiplication. Let me mention right away the two basic sources of nice multiplicative sets S in R: (1) Let f S. Then S = {1, f, f 2,... } is a multiplicative set generated by f; (2) Let P be a prime ideal of R. Then, S = R P (the complement of P in R) is a multiplicative set by the definition of prime ideal. Now let S be any multiplicative set in R. Define an equivalence relation on R S by (r, s) (r, s ) if and only if (rs r s)t = 0 for some t S. Write S 1 R for the set of equivalence classes in R S, and write r/s S 1 R for the equivalence class of an element (r, s) R S. We wish to give S 1 R the structure of a ring, so that the map i S : R S 1 R, r r/1 is a ring homomorphism. To do this, take r/s, r /s S 1 R and define r/s + r /s = (rs + r s)/(ss ), (r/s)(r /s ) = (rr )/(ss ). Of course, we need to check these formulae are well-defined. For example, for the first, suppose r /s = r /s. Then, there exists t S such that (r s s r )t = 0. We need to show for well-definedness that (rs + r s)/(ss ) = (rs + r s)/(ss ). By definition, this means that we need to find a u S such that (rs + r s)ss u (rs + r s)ss u = r sss u r sss u = 0. But u = t works! Similarly, one checks that the other formula is well-defined. Then, these formulae give on S 1 R the addition and multiplication making it into a commutative ring. This ring S 1 R, together with the ring homomorphism i S : R S 1 R, r r/1 is called the localization of R at S. Note that in S 1 R, each element i S (s) for s S is a unit: the inverse element is 1/s. Localization is characterized by the following universal property: Universal property of localization. Given a homomorphism f : R R to a commutative ring R such that f(s) is a unit for each s S, there exists a unique ring homomorphism f : S 1 R R such that f = f i S R. Proof. There is no choice but to define f(r/s) = f(r)f(s) 1. You just need to check that this is well-defined! It is reasonable to ask when the map i S : R S 1 R is injective: Lemma. i S is injective if and only if S contains no zero divisors. (Remember: a zero divisor is a z S such that zy = 0 for some y S. Thus, zero is a zero divisor except in the zero ring!)

10 50 CHAPTER 2. RINGS Proof. Suppose that i S (r) = 0 for r R. Then, r/1 = 0/1, i.e. there exists s S such that rs = 0, i.e. s is a zero divisor. So if the set S has no zero divisors, you can regard R as a subring of S 1 R and think of the construction of S 1 R as adjoining inverses to the elements of S to make them into units. At the other extreme, it is perfectly possible for S 1 R to be the zero ring: suppose there exist elements s, t S such that st = 0 (for instance if S contains a nilpotent element). Then, we have that rst = 0 hence r/1 = rs/s = 0/1 for every r R, so that S 1 R = {0/1}. For some simple examples of localization, start with the ring Z and a prime p. Then: (1) S = {1, p, p 2,... } is a multiplicative set and we can identify the localization S 1 Z with the set of all rational numbers of the form {n/p r n Z, r 0} (proof: check this ring satisfies the universal property of localization). (2) Now let P = (p), a prime ideal in Z, and let S = Z P be the multiplicative set arising as its complement. In this case, the localization S 1 Z can be identified with the set of all rational numbers of the form {n/m n Z, m N, (m, p) = 1}. One of the most important application of localization is to construct the field of fractions of an integral domain R. So suppose now that R is an integral domain, i.e. that (0) is a prime ideal. Then, S = R is a multiplicative set containing no zero divisors. So, applying the lemma, there is an injective ring homomorphism i : R Q, where Q is the the localization of R at R. The ring Q, together with the embedding i : R Q, is called the field of fractions of R: it is a field since every element of R is invertible in Q. Note elements of Q look like r/s for r R, s R and addition and multiplication is just by the usual laws of fractions. Let us record the universal property characterizing the pair (Q, i), which is a special case of the universal property of localization. Universal property of fractions. Given an integral domain R, there is a field Q and a monomorphism i : R Q such that for any other field Q and monomorphism f : R Q, there is a unique monomorphism f : Q Q such that f = f i. Note the field of fractions Q of the integral domain R is uniquely determined (up to canonical isomorphism) by this universal property. The universal property says that the field of fractions R is the smallest field that R can be embedded into as a subring. For some examples: (1) The field of fractions of Z is the field Q of all rational numbers. (2) The field of fractions of F [X], X an indeterminate and F a field, is the field of rational functions in the indeterminate X, denoted F (X). Elements of F (X) look like fractions f(x)/g(x) for polynomials f, g F [X] with g 0. (3) More generally, the field of fractions of F [X i i I] where the X i indeterminates, is the field F (X i i I) of rational functions in indeterminates X i. 2.6 Factorization in polynomial rings I want to prove in this section that if R is a UFD then the polynomial ring R[X] in an indeterminate X is also a UFD. Hence, for example, the polynomial ring F [X 1,..., X n ] in indeterminates over a field F is a UFD (because fields are vacuously UFDs). On the other hand, F [X 1,..., X n ] is not a PID if n 2, so this gives an example of a PID that is not a UFD. You should recall the division algorithm in a polynomial ring R[X] explained in section 2.3. We will also make use of the basic notion of localization from section 2.5. Now let R be a UFD for the remainder of the section. Given f R[X] with f = a n X n + + a 1 X + a 0 we define its content to be the greatest common divisor C(f) = GCD(a 0, a 1,..., a n ). Of course, C(f) is only defined up to associates. Call f primitive if its content is a unit. For 0 f R[X] there exists a primitive f R[X] such that f = C(f)f.

11 2.6. FACTORIZATION IN POLYNOMIAL RINGS 51 Gauss lemma. Given f, g R[X], both f and g are primitive if and only if fg is primitive. Proof. Every coefficient of fg is divisible by the GCD of the coefficients of f and of g. So if fg is primitive, so are both f and g. Conversely, take f = a n X n + + a 1 X + a 0, g = b m X m + + b 1 X + b 0 both primitive. We need to show that fg is primitive. Well, for each prime p R, there is at least one coefficient a s of f with p a s and one coefficient b t of g with p b t. Take s and t to be the minimal such indices. Consider the coefficient of X s+t in fg, namely c = a i b j. i+j=s+t By choice of s and t, p a i for i < s and p b j for j < t. Hence, p a i b j for all i + j = s + t except for a s b t. Since p is irreducible, p a s b t. Hence, p c, so fg is primitive. Let Q be the field of fractions of R. Since R is an integral domain, we can view R as a subring of Q in the natural way. In turn, we can view R[X] as a subring of Q[X]. Now, R is a UFD and Q[X] is a PID hence a UFD Lemma. Let f, g R[X] be primitive. Then: (i) f and g are associates in R[X] if and only if they are associates in Q[X]; (ii) f is irreducible in R[X] if and only if it is irreducible in F [X]. Proof. (i) One direction is obvious! Conversely, suppose that f and g are associates in Q[X]. This implies that af = bg for some a, b R. Now, let p be a prime and suppose p n a. Then, p n divides af hence p n divides bg. Since p is prime and g is primitive, this implies p n divides b. Now you deduce since R is a UFD that a and b are associates in R. Hence, a = bu for a unit u R, whence af = bfu = bg whence fu = g since R[X] is an integral domain. Hence, f and g are associates in R. (ii) Suppose f is reducible in F [X]. Then, f = g 1 h 1 for g 1, h 1 F [X] with deg g 1, deg h 1 > 0. Clearing denominators, we can find some d R such that df = g 2 h 2 with g 2, h 2 R[X] of degree > 0. So there exist primitive polynomials g, h R[X] of degree > 0 such that df = cgh where c = C(g 2 )C(h 2 ). By Gauss lemma, gh is primitive, and f is primitive by assumption. We deduce that p n d if and only if p n c for p R prime. Hence, since R is a UFD, c and d are associates, so c = ud for u R a unit. Hence, f = ugh in R[X], i.e. f is reducible in R[X]. Now we can prove the main result: Theorem. If R is a UFD, then R[X] is a UFD. Proof. Let f R[X] be a non-zero non-unit. Write f = C(f)f for a primitive polynomial f. Since R is a UFD, C(f) is either a unit or has unique factorization in R hence in R[X] (its a scalar!). In other words, we can reduce right away to the case C(f) is a unit, i.e. f = f is primitive. Now since Q[X] is a UFD, there exist irreducible g 1,..., g n Q[X] with f = g 1... g n in Q[X]. Pick primitive f i R[X] and d i, c i R so that d i g i = c i f i for each i. Since g i is irreducible in Q[X], so is f i, hence by (ii) of the lemma, f i is irreducible in R[X] too. So with d = d 1... d n, c = c 1,..., c n, we have shown df = cf 1... f n

12 52 CHAPTER 2. RINGS where each f i is primitive and irreducible. Then by Gauss lemma, f 1,..., f n is primitive. Hence, for any prime p in R, p n d if and only if p n df if and only if p n cf 1... f n if and only if p n c. This shows d and c are associates in R. Cancelling, we deduce that f = f 1... f n for irreducibles f 1,..., f n R[X]. This gives existence of factorization in R[X]. Now for uniqueness, suppose instead that f = g 1... g m for irreducibles g 1,..., g m R[X]. Since f is primitive, Gauss lemma gives that each f i and g j is too. So by (ii) of the lemma, f i and g j are irreducible in Q[X]. So since Q[X] is a UFD, we get that m = n and (after reordering) f i and g i are associates in Q[X]. By (i) of the lemma, this gives that f i and g i are associates in R[X] too, as required. Now seems like a good place to also include the useful: Eisenstein s criterion. Let R be a UFD and f = a n X n + + a 1 X + a 0 R[X] by primitive. If there is some prime p R such that p a n, p a i for i = 0,..., n 1 and p 2 a 0, then f is irreducible in R[X]. Proof. Suppose to the contrary that we can write f = gh in R[X] where deg g, deg h > 0. Say g = b m X m + + b 1 X + b 0, h = c k X k + + c 1 X + c 0 where b m, c k 0. Since a n = b m c k we see that p divides neither b m nor c k. Since p a 0 = b 0 c 0 and p 2 a 0, we see that p divides exactly one of b 0, c 0, say p b 0, p c 0. Now take 0 < j m minimal such that p b j 1 but p b j. The coefficient of X j in f is a j = b j c 0 + b j 1 c b 0 c j Since m < n, p divides a j, hence divides the right hand side, hence divides b j c 0, a contradiction. For example, the polynomial X 3 + 3X 2 + 3X + 3 is irreducible in Z[X] by Eisenstein for p = 3. Since it is monic, hence primitive, it follows by Gauss lemma as above that it is irreducible in Q[X] too. Other than this handy test (and variations on it obtained by substitutions of the form X = p(y ) for a polynomial p, see example below), it can in general by very hard to tell if a polynomial in R[X] is irreducible or not. Here is one more important example. Let n be a positive integer. The nth cyclotomic polynomial is defined from Φ n (x) = (x ω) where the product is over all primitive nth roots of unity ω C. Of course, the degree of the polynomial Φ n (x) is φ(n), where φ is Euler s φ-function. Note that x n 1 = d n Φ d (x). Using this and induction on n, you deduce that Φ n (x) is actually a monic polynomial in Z[x]. In fact Φ n (x) is irreducible. The proof of this takes a little work so I m not going to do it here. But there is one trivial case: if n = p is prime then Φ p (x) = x p 1 + x p x 2 + x + 1 = xp 1 x 1. Let me prove using Eisenstein s criterion that this is irreducible. Subsitute x = y + 1. You obtain (y + 1) p ( ) ( ) ( ) 1 p p p = y p 1 + y p y +. y 1 p 2 p 1 All the coefficients except the leading coefficient are divisible by p, and the constant term is not divisible by p 2. Hence by Eisenstein, this polynomial is irreducible. Hence, the original polynomial x p x + 1 must also be irreducible.

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