Unique Factorization

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1 Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon of unique factorization in rings, as a generalization of unique factorization of integers into primes. First, we ll define what unique factorization means in general. Definition 1.1. A (commutative) ring R is a domain (or integral domain) if whenever ab = 0 for a, b R, then either a = 0 or b = 0. An element u R is a unit if it has an inverse, i.e. if there exists u 1 R such that uu 1 = 1. If every nonzero element of R is a unit, we say R is a field. Domains will be the basic setting in which it will make sense to ask questions about unique factorization. Units are important because they are the ambiguity in unique factorizations. For example, in Z, the only units are 1 and 1. The fact that 1 is a unit means that factorization of integers into primes isn t really unique. For example, we can factorize 6 = 2 3 or 6 = ( 2) ( 3). However, we still consider these two factorizations to be the same since 2 and 2 (respectively 3 and 3) only differ by multiplication by the unit 1. Definition 1.2. Let R be a domain, and a, b R. We write a b (a divides b, or a is a factor of b) if there exists c such that b = ac. We say a nonzero element p R is irreducible if it is not a unit and if whenever a p, either a is a unit or a = up for some unit u. Alternatively, whenever we can write p = ab, one of a and b must be a unit. a. Note that any factor of a unit is again a unit. Indeed, if u = ab, then abu 1 = 1 so bu 1 is an inverse for We finally define unique factorization itself. Definition 1.3. A domain R has unique factorization (or is a unique factorization domain UFD) if every nonzero a R can be written uniquely as a product a = u p di i where u is a unit and the p i are irreducible. Here uniquely means up to reordering the factors and multiplying the factors by units. Whew! That s a lot of definitions. Let s look at some examples. Example 1.4. The integers Z are a domain but not a field. The only units in Z are 1 and 1. An integer is irreducible iff it is either a prime number or negative a prime number. By the fundamental theorem of arithmetic (which we will prove later), Z is a UFD. Example 1.5. The rationals Q, the reals R, and the complex numbers C are all fields. Any field has no irreducibles, and is automatically a UFD 1

2 Example 1.6. Any subring of a domain is a domain. In particular, any collection of complex numbers closed under addition, subtraction, and multiplication is a domain. For example, for any integer n Z, there is a domain Z[ n] = {a + b n : a, b Z} since it can be checked that such numbers are closed under addition, subtraction, and multiplication. Example 1.7. Let s look more closely at a special case of the example above, namely Z[ 1] = Z[i], also known as the Gaussian integers. The Gaussian integers are a lot like the integers, but some things work a little differently. First of all, besides 1 and 1, i and i are also units since i ( i) = 1. Also, for instance, 2 is irreducible as an integer, but not as a Gaussian integer, since we can factor it as 2 = (1 + i)(1 i). More generally, if p is a prime which can be written as a sum of two squares p = a 2 + b 2, then p is not irreducible as a Gaussian integer because it factors p = (a + bi)(a bi). To say much more about factorization in Z[i], we need to use the norm N(a + bi) = a + bi 2 = a 2 + b 2, a map from Z[i] to the nonnegative integers. The key property of the norm is that N(xy) = N(x)N(y). In particular, if u is a unit, then 1 = N(1) = N(uu 1 ) = N(u)N(u 1 ), which implies N(u) must be 1. It follows that ±1 and ±i are the only units in Z[i]. Similarly, let p be a prime, and suppose that in Z[i], we can factor p = xy in a nontrivial way. Then N(p) = p 2 = N(x)N(y). Since x and y are not units, we must have N(x) = N(y) = p, so p must be a sum of two squares (say if x = a + bi, then p = a 2 + b 2 ). Furthermore, the same argument shows that x and y themselves must be irreducible. On the other hand, if p cannot be written as a sum of two squares, it is irreducible in Z[i]. We have now figured out how to factor any ordinary integer n into irreducible Gaussian integers. Namely, for each ordinary prime factor p, if p is not a sum of two squares, it is irreducible, and if p = a 2 + b 2, then we can replace p in the factorization of n with (a + bi)(a bi), and each of these is irreducible. It turns out that Z[i] is a UFD; we will prove this later. As a consequence, we can show that in fact the above tells us everything about factorization in Z[i]: the irreducibles of Z[i] are exactly the primes p which are not sums of two squares and a ± bi such that a 2 + b 2 is prime. To show this, let x = a + bi Z[i]. Then N(x) = a 2 + b 2 = x x. We know that all the irreducible factors of N(x) are of the form mentioned above, since N(x) is an ordinary integer. By unique factorization, every irreducible factor of x is also one of N(x). Thus every irreducible factor of x must be of the form above, and so every irreducible Gaussian integer is of the form above. Example 1.8. Almost everything in the discussion above also applies to the example Z[ 5] (or of Z[ n] for any n). The only difference is that the norm is given by N(a + b 5) = a 2 + 5b 2, so everywhere we said sum of two squares we should now say of the form a 2 + 5b 2. However, there is one big difference between Z[ 5] and Z[i]: Z[ 5] is not a UFD! To see this, let s factor 9 in Z[ 5]. Well, first of all 9 = 3 3, and 3 is irreducible since it is not of the form a 2 + 5b 2. On the other hand, 9 = , so 9 = (2 + 5)(2 5). Note that 2 ± 5 is irreducible for the same reason 3 is: it has norm 9, and no factor can have norm 3 since 3 is not of the form a 2 + 5b 2. Clearly 3 does not differ from 2 ± 5 by a unit, so we conclude that 9 has two different factorizations in this ring! 1.1 Exercises Exercise 1.1. Factor i into irreducible Gaussian integers. (Hint: Use the norm.) Exercise 1.2. Let R be a UFD and let p R be irreducible. Show that if p ab for some a, b R, then either p a or p b. (Tomorrow, we will show that this property is basically equivalent to R being a UFD). Exercise 1.3. Show that Z[2i] = {a + bi : a, b Z and b is even} is not a UFD. Exercise 1.4. Show that Z[ 6] is not a UFD. (Hint: Find an example similar to the example for Z[ 5].) 2

3 Exercise 1.5. Let k be a field and let R = k[x, y, z, w]/(xy zw), the quotient of the polynomial ring k[x, y, z, w] in four variables by the ideal generated by xy zw. Assuming that R is a domain (which we will prove later), show that it does not have unique factorization. (Hint: xy = zw looks like two different factorizations of the same element. Can you show that it really is?) Exercise 1.6. Let R be a domain and a, b R be nonzero. Then a greatest common divisor (or gcd) of a and b is a d R such that d a, d b, and whenever c a and c b, c d. (a): Show that the gcd of two elements, if it exists, is unique up to multiplication by a unit. (b): Suppose R is a UFD. Show that any two elements have a gcd. 3

5 Definition 2.5. A nonzero nonunit element p of a domain is prime if whenever p ab, p a or p b. Note that primes are automatically irreducible: indeed, if p factors as ab, then p ab, so (say) p a and a = pc Then p = ab = pbc, so bc = 1, so b is a unit. Furthermore, in a UFD, irreducibles are prime: if p ab, a factorization of ab is given by combining factorizations of a and b, so p must divide one of them by uniqueness. In fact, this is actually equivalent to unique factorization! Theorem 2.6. Let R be a domain such that every nonzero a R has a factorization into irreducible elements. Then if every irreducible element of R is prime, R is a UFD. Proof. The proof is exactly the same as the proof above in the case Z = R. Thus the hard work is really in showing that irreducibles are prime, i.e. Lemma We now give a proof of this in the case of R = Z that will generalize to many other cases. We first need a definition. Definition 2.7. Recall that an ideal in a ring R is a subset I R closed under addition such that for any a R and i I, ai I. For any a R, we have an ideal (a) = {ar : r R} = {b : a b}; we call such ideals principal. If every ideal in a domain R is principal, we say R is a principal ideal domain (PID). Here is one concrete consequence of being a PID. Let a, b R. Then the ideal (a, b) = {ar + bs : r, s R} must be principal; say (a, b) = (c). Then a, b (c), so c a and c b. Furthermore, we can write c = ar + bs for some r and s, which implies that every other common factor of a and b also divides c. That is, c is a greatest common divisor (gcd) of a and b, and is a combination ar + bs of a and b. Theorem 2.8. In a PID, irreducibles are prime. Proof. Let R be a PID and p R be irreducible. Suppose p ab and p b; we will show p a. Consider the ideal (p, b) = (c), where c is a gcd of p and b. Then c p, so since p is irreducible, either c is a unit or c is the same as p, up to a unit. If c is the same as p, then (p, b) = (c) = (p) so b (p) and p b, contradicting our assumption. Thus c is a unit, and so we may assume c = 1. We can thus write 1 = rp + sb for some r and s. Multiplying this equation by a, we get a = rpa + sba. But p ab, so p divides the right-hand side, so p a, as desired. To complete the proof of the fundamental theorem of arithmetic, we thus must only prove: Theorem 2.9. Z is a PID. Proof. Let I Z be an ideal. If I = {0}, then I = (0) is principal. Otherwise, I contains some nonzero n, and by multiplying by ±1, I contains some n > 0. Let n be the smallest positive element of I. Then I (n). For any m I, by division with remainder, we can write m = qn + r where 0 r < n. Since m I and qn I, it follows that r I. But r < n, so by minimality of n we must have r = 0. Thus m = qn is divisible by n, so I = (n). We now have a complete proof that Z is a UFD. More importantly, however, we have a machine set up to show that other domains are also UFDs: we just have to show they are PIDs in which factorizations always exist. Tomorrow, we will use this to prove that other rings are UFDs. 2.1 Exercises Exercise 2.1. Let R be a domain and a R be irreducible. Show that a is also irreducible as an element of the polynomial ring R[x]. Exercise 2.2. Let k[x 2, x 3 ] k[x] be the ring of polynomials with coefficients in a field k with no linear term (which is the subring generated by k, x 2, and x 3 ). Show that k[x 2, x 3 ] is not a UFD. (Hint: Show that x 2 and x 3 are irreducible.) 2 We also need to show that factorizations exist at all, but this is usually easy by some sort of induction argument, as we gave for Z. 5

6 Exercise 2.3. Let k be a field. Show that in k[x], every nonzero element can be factored (maybe not uniquely) as u p di i where u is a unit and the p i are irreducible. (Hint: Use induction on the degree of the polynomial.) Exercise 2.4. Let k be a field. Can you prove that k[x] is a PID by imitating the proof for Z? (We will talk about this in class tomorrow.) Exercise 2.5. Let k be a field. Show that in k[x 1,..., x n ], every nonzero element can be factored (maybe not uniquely) as u p di i where u is a unit and the p i are irreducible. (Hint: Define a notion of degree for a polynomial in multiple variables.) Exercise 2.6. An ideal I R in a ring R is called prime if the quotient ring R/I is a domain. (a): Show that I is prime iff whenever ab I, either a I or b I. (b): Show that an element a R is prime iff the ideal (a) is a prime ideal. Exercise 2.7. An ideal I R in a ring is called maximal if the quotient ring R/I is a field. (a): Show that I is maximal iff any ideal J I containing I is equal to either I or R. (Hint: Show that a ring is a field iff it has no ideals other than {0} and the whole ring.) (b): Show that maximal ideals are prime. (c): Show that in a PID, prime ideals (other than {0}) are maximal. (Hint: Note that (a) (b) iff b a.) Exercise 2.8 (Chinese remainder theorem). Let R be a PID, and let a, b R with gcd(a, b) = 1. Define a map f : R/(ab) R/(a) R/(b) by f(x + (ab)) = (x + (a), x + (b)). Show that f is a ring isomorphism. 3 (For example, for R = Z, this says that if m and n are relatively prime, then Z/mn = Z/m Z/n.) 3 If R and S is a ring, then R S is also a ring by adding and multiplying separately on each coordinate. In particular, R/(a) R/(b) is a ring. 6

7 3 Euclidean domains The argument we made yesterday for Z readily generalizes to some other domains. Theorem 3.1. Let k be a field. Then k[x] is a PID. Proof. We do the same thing as we did for Z, except instead of taking a minimal positive element, we take an element of minimal degree. More precisely, let I k[x] be an ideal, which we can assume to contain some nonzero element. Let f(x) I be nonzero of minimal degree and let g(x) I; we will show that f g and hence I = (f). By polynomial long division, we can divide g = qf + r where the remainder r has degree strictly less than the degree of f. Since g I and qf I, we also have r I. But by minimality of the degree of f, this can only happen if r = 0. Thus g = qf so I = (f). Corollary 3.2. Let k be a field. Then k[x] is a UFD. Proof. We only need the existence of factorizations. This can be proven by induction on degree, or using the fact that we will prove later that in fact PIDs always have factorizations and hence are automatically UFDs. We now make a definition which generalizes the arguments above for Z and k[x]. Definition 3.3. A Euclidean domain is a domain R equipped with a function d : R {0} N such that: For any nonzero a, b R, d(ab) max(d(a), d(b)), and For any nonzero a, b R, we can write a = bq + r where either r = 0 or d(r) < d(b). Example 3.4. Z is a Euclidean domain with d(n) = n. Example 3.5. k[x] is a Euclidean domain with d(f) = deg(f). Theorem 3.6. Euclidean domains are PIDs (and hence UFDs). Proof. Let R be a Euclidean domain. The proof that R is a PID is exactly the same as the proof for R = Z or R = k[x]: given an ideal I, take a I such that d(a) is minimized, and then a will generate I. We can finally give an application of all this machinery: a quick proof that Z[i] is a UFD. Theorem 3.7. Z[i] is a Euclidean domain with d(a + bi) = N(a + bi) = a 2 + b 2. Proof. The first condition on d is easy to check. For the second condition, let x, y Z[i]; we wish to divide y by x with a remainder which has smaller norm than x. That is, we want to find a multiple of x which is closer to y than x is to 0. For this, we use the geometry of Gaussian integers in the complex plane: the multiples of x form a square lattice, and elementary geometry shows that any point in the plane is closer to some vertex of the lattice than the side length of the squares. Corollary 3.8. Z[i] is a UFD. This proof also can be used to see what goes wrong for Z[ 5]. In that case, instead of having a square lattice we have a rectangular lattice, and the geometric fact we used is no longer true. In the previous section, we made a geometric argument that the norm made Z[i] a Euclidean domain, but this argument failed for Z[ 5]. Let s recall that argument: the point is that to get division with a smaller remainder, you need that in a square with side length 1, every point is distance < 1 from some corner. In the case of Z[ 5], instead of a square we had a rectangle, with one side having length 1 and the other side having length 5. More generally, if we had done Z[ d] for any d > 0, the other side would have length d. A little geometry shows that the maximum distance a point can be from a corner is d + 1/2. Thus this argument still works for d = 2 (Exercise 3.2), breaks down for d > 3, and just barely fails for d = 3. We now turn to the general question of the existence of factorizations in PIDs. 7

8 Lemma 3.9. Let R be a PID and a R be a non-unit. Then a has an irreducible factor. Proof. Suppose a = a 0 has no irreducible factor. Then we can pick a nontrivial factor a 1 a 0, and a 1 cannot have an irreducible factor either. Thus we can pick an infinite sequence... a 2 a 1 a 0, where each a n is a nontrivial factor of a n 1. This means exactly that (a 0 ) (a 1 ) (a 2 )... with each inclusion being strict. Now consider the ideal I = (a 0, a 1,... ) generated by all the a i. Since R is a PID, I is generated by some single element c. This c can be written as a finite linear combination of the a i, say c = n b i a i. But then c (a n ) since a n a i for every i n. But then we have (a n ) I = (c) (a n ), which is a contradiction. Hence no such a = a 0 could have existed, so any element has an irreducible factor. Theorem Let R be a PID. Then every nonzero element a R has a factorization into irreducible factors (and a unit). Proof. We may assume a = a 0 is not a unit. Thus by Lemma 3.9, a has some irreducible factor p 1 ; say a = p 1 a 1. If a 1 is a unit we re done, otherwise, we can write a 1 = p 2 a 2 for some irreducible p 2. Continuing by induction, if we never get a unit, we get an infinite sequence... a 2 a 1 a 0 of nontrivial factors, which gives a contradiction as in the proof of Lemma 3.9. Thus some a n is eventually a unit, so we get a factorization a = a n p 1 p n. Corollary Let R be a PID. Then R is a UFD. 3.1 Exercises Exercise 3.1. Let k be a field. For f(x, y) k[x, y], let deg(f) be the largest degree of a term of f, where the degree of a monomial x m y n is defined to be m + n. Show that deg does not make k[x, y] a Euclidean domain. Exercise 3.2. Show that Z[ 2] is a Euclidean domain, and thus a PID and a UFD. (Hint: Imitate the proof for Z[i].) I don t know how to do the following exercise, but I know that it s true. I d love to see a solution if you can find one. The difficulty is that you can t use the geometry of complex numbers, so we can t make exactly the same sort of argument as we did for Z[i]. Exercise 3.3. Show that Z[ 2] is a Euclidean domain. (Hint: I think d(a + b 2) = a 2 + 2b 2 works.) Exercise 3.4. Call a Euclidean domain R strictly Euclidean if whenever neither a nor b is a unit, then d(ab) > max(d(a), d(b)). For example, Z, k[x], and Z[i] are all strictly Euclidean. Show using induction on d(a) (instead of the argument using ideals given in class) that if R is strictly Euclidean, then any nonzero a R can be factored into irreducibles. Exercise 3.5. Let R be a domain. Then the field of fractions K(R) of R is the collection of expressions a/b with a, b R, b 0, where we say a/b = c/d if ad = bc. We make K(R) a ring using the ordinary rules for addition and multiplication of fractions. (a): Show that by mapping a R to a/1, we can think of R as a subring of K. (b): Suppose R is a UFD. Show that any nonzero element of K(R) can be factored as u p di i where u R is a unit, p i R are irreducible, and d i Z are possibly negative, and this factorization is unique up to changing the factors by units in R. 8

9 4 Factorization in polynomial rings In the previous section, we showed that if k is a field, then the ring of polynomials k[x] is a UFD. In this section we will look at polynomials with more general coefficients. Our basic example will be Z[x], the ring of polynomials with integer coefficients. We can start by trying to use the same method we did in the case of a field: we want to show that Z[x] is a Euclidean domain with d(f) = deg(f). The main thing to check is that given any f and g, we can write g = qf + r where the remainder r has lower degree than f. Over a field, we can do this by polynomial long division. However, over Z this breaks down. Consider, for example, g(x) = x 2 +x+2 and f(x) = 2x+1. Performing long division would give quotient 1 2 x with remainder 7/4. However, we re not allowed to do this since the coefficients must remain integers. If the quotient has to have integer coefficients, there s no way we can use a multiple of f to cancel out the x 2 and x terms in g and leave an integer remainder. Indeed, it is impossible to make Z[x] a Euclidean domain since it is not a PID! For example, the ideal (2, x) is not principal: it is the set of polynomials with even constant term, which is not just the multiples of any one polynomial. Alternatively, although clearly the gcd of 2 and x should be 1, there is no way to write 1 as a linear combination of 2 and x. Nevertheless, it turns out that Z[x] is a UFD. To show this, we will need a new argument different from the ones we ve used. The basic idea will be to use the fact that both Z and Q[x] are UFDs, and the fact that Z[x] is kind of halfway between them. Let s look at what irreducibles in Z[x] look like. There are two kinds, namely the primes in Z as constant polynomials, and the irreducible polynomials of degree > 0. In particular, note that if f(x) Z[x] is irreducible as an element of Q[x] and monic, it is also automatically irreducible in Z[x], since there are only fewer possible factorizations in Z[x] (and since it is monic, there is no issue with the fact that Z[x] has fewer units). Thus the first kind of irreducibles are the irreducibles of Z, and the second kind at least include some irreducibles coming from Q[x]. To show Z[x] is a UFD, the main thing we need to show is that all irreducibles are prime, so we will start by showing this for the irreducibles coming from Z. In fact, nothing we will do is special to Z, so we will replace it with an arbitrary UFD. Lemma 4.1 (Gauss s Lemma). Let R be a domain and let p R be prime (e.g., R is a UFD and p is irreducible). Then p is also prime as an element of R[x]. Proof. Let s see what it means for p to be prime in R[x]. Note that p divides a polynomial iff it divides each of its coefficients. Thus the claim is that if p divides all the coefficients of fg, it must divide either all the coefficients of f or all the coefficients of g. This is far from obvious, since the coefficients of fg are complicated expressions in the coefficients of f and g. Let s look at an example: say f(x) = ax + b and g(x) = cx + d. Then f(x)g(x) = acx 2 + (ad + bc)x + bd. Let s suppose all the coefficients of the product are divisible by p. Then in particular, p ac, so p a or p c. Say p a (the other case is similar). Then since p ad + bc, the next coefficient, we must have p bc so p b or p c. If p b, then p ax + b = f and we re done. If p b, then p c, and also since p bd, the next coefficient, p d. Thus p cx + d = g, and we re done again. The proof for general polynomials is similar to this example. Say deg(f) = n and deg(g) = m and write and f(x) = a n x n + + a 0 g(x) = b m x m + + b 0. Let s assume that p f; we will show p g. There is some coefficient of f that p does not divide; let a r be the largest one (i.e. the one with r maximal). We show by descending induction that p divides each coefficient 9

10 of g. First, the coefficient of x m+r in f(x)g(x) will be b m a r + b m 1 a r+1 + b m 2 a r By maximality of r, p divides a r+1, a r+2, and so on, so p divides every term but the first. But p fg so p must divide the entire coefficient, so p b m a r. Since p a r, p b m. Now let k < m and suppose p b l for all l > k. Then the coefficient of x k+r in f(x)g(x) is + b k+1 a r 1 + b k a r + b k 1 a r Since p b l for l > k and p a s for s > r, we know that p divides every term of this except b k a r. Thus p b k a r and so p b k. Gauss s lemma not only tells us that the irreducibles from Z are still prime in Z[x]; it also tells us something about the nonconstant irreducibles. For this, we first need a definition. Definition 4.2. Let R be a UFD and f R[x]. Then the content cont(f) of f (defined up to a unit) is the gcd of all the coefficients of f (gcd makes sense because R is a UFD). If cont(f) = 1, we say f is primitive. Note that for any f, we can factor f = cont(f)f 1 where f 1 is primitive. Corollary 4.3. Let R be a UFD and f, g R[x]. Then cont(fg) = cont(f) cont(g). Proof. By factoring f = cont(f)f 1 and g = cont(g)g 1, we get fg = cont(f) cont(g)f 1 g 1. It thus suffices to show that f 1 g 1 is primitive. If it were not primitive, some irreducible p R would divide it. But by Gauss s Lemma, this would imply p f 1 or p g 1, which is impossible since they are primitive. Corollary 4.4. A nonconstant element of Z[x] is irreducible iff it is primitive and is irreducible as an element of Q[x]. Proof. First, suppose f Z[x] is irreducible and nonconstant. It clearly must be primitive. Suppose we can factor f = gh for g, h Q[x]. Note that content also makese sense for elements of Q[x]: it is the unique rational cont(g) (up to a unit in Z such that g = cont(g)g 1 for some primitive g 1 Z[x] (explicitly, write all the coefficients of g in lowest terms, and take the gcd of the numerators divided by the lcm of the denominators). We can thus write f = cont(g) cont(h)g 1 h 1, where now g 1, h 1 Z[x] are primitive and cont(g), cont(h) Q. By Corollary 4.3, g 1 h 1 is primitive as well. But f is primitive, so this implies cont(g) cont(h) must be a unit in Z. But then, up to a unit, we can factor f = g 1 h 1 in Z[x]. Since f is irreducible in Z[x], this implies g 1 or h 1 is a unit, so then g or h is a unit in Q[x]. Thus f is irreducible in Q[x]. Conversely, suppose f is primitive and is irreducible as an element of Q[x], and suppose we can factor f = gh in Z[x]. Since f is irreducible in Q[x], g or h must be a unit in Q[x], i.e. an element of Q (or actually Z since it is also in Z[x]). But since f is primitive, f has no nontrivial factors in Z, so g or h must be a unit in Z. Thus f is irreducible in Z[x]. Using this, we can finally prove: Theorem 4.5. Z[x] is a UFD. Proof. First, we show the existence of factorizations. Let f Z[x] and factor f = q i where the q i are irreducible elements of Q[x] (this is possible since Q[x] is a UFD). By factoring out the contents of the q i, we can rewrite this as f = c q i, where q i Z[x] is primitive and irreducible in Q[x] and c Q is the product of the contents of the q i. By Corollary 4.4, the q i are irreducible in Z[x], and by Corollary 4.3, c is equal to the content of f and is thus actually in Z. Factoring c into primes p j in Z then gives a factorization f = p j q i of f into irreducibles of Z[x]. We now show that every irreducible in Z[x] is prime. By Gauss s lemma, it suffices to show this for nonconstant irreducibles f. Suppose f gh. By Corollary 4.4, f is irreducible in Q[x] and hence prime in Q[x] 10

11 since Q[x] is a UFD. Thus in Q[x], f g or f h; say f g. We can then write g = qf, where q Q[x]. Now we can write q = cont(q)q 1 where cont(q) Q and q 1 Z[x] is primitive. Since f is primitive, it follows that cont(q) = cont(g) is actually an integer, so in fact q = cont(q)q 1 is actually in Z[x]. Thus f g in Z[x]. and f is prime. Finally, we observe that in all of this, there was nothing special about Z: we could have replaced Z by any UFD R, as long as we also replace Q by the field of fractions of R. Recall that the field of fractions of a domain R is the set of formal fractions a/b with a, b R, b 0; this is a field with the usual multiplication and addition of fractions and R embeds in it by sending a to a/1. Theorem 4.6. Let R be a UFD. Then R[x] is a UFD. Proof. Exactly the same as the proof of Corollary 4.4 and Theorem 4.5, with Z replaced by R and Q replaced by its field of fractions. In particular, by induction we get that a polynomial ring k[x, y, z,... ] or Z[x, y, z,... ] in any (finite) number of variables is a UFD. Furthermore, we can use Corollary 4.4 to get an inductive criterion for detecting irreducibility of elements of these polynomial rings. However, this criterion is hard to use since the fields of fractions that come up in the induction steps, such as the field of fractions of k[x, y], are not particularly easy to work with. One useful application of unique factorization is showing the quotients are domains. By Exercise 2.6, if R is a domain and p R is prime, then the quotient R/(p) is a domain. In general, it can be quite difficult to show that a particular ring defined as a quotient is a domain. However, for UFDs, we get that R/(p) is a domain for any irreducible p. In particular, whenever you have an irreducible polynomial f k[x, y, z,... ], the quotient k[x, y, z,... ]/(f) is a domain. 4.1 Exercises Exercise 4.1 (Eisenstein s criterion). Let R be a UFD, p R be prime, and let f(x) = a n x n + a n 1 x n 1 + +a 0 R[x] be a polynomial. Suppose that p a i for all i < n, p a n and p 2 a 0. Show that f is irreducible. Exercise 4.2. Let k be a field. Show that k[x, y, z, w]/(xy zw) is a domain but not a UFD. (Hint: See Exercises 1.5 and 2.6.) Exercise 4.3. Let R be a ring that is not a field. Show that R[x] is not a PID. Exercise 4.4. Let R be a UFD and suppose d R is not a square (i.e., d a 2 for any a R). (a): Show that x 2 d R[x] is irreducible. (b): Suppose R = Z. Show that Z[x]/(x 2 d) is isomorphic to Z[ d]. (This gives a new proof that Z[ d] is a domain, which we originally showed by saying it is a subring of C.) 11

12 5 Dedekind domains Let s look more closely at the case Z[ 3]. By trial and error, we can find that unique factorization fails: for example, 2 2 = (1 + 3)(1 3). However, it turns out that in this case there is a more general reason why unique factorization must fail. To see this, we first need some definitions. Definition 5.1. Let S be a ring and let R S be a subring, and let a S. We say a is integral over R if there is a monic polynomial f(x) = x n + c n 1 x n c 0 R[x] such that f(a) = 0. We say a domain R is integrally closed if whenever a K(R), the field of fractions of R, is integral over R, then in fact a R. Note that, for example, any a R is integral over R, by letting f(x) = x a. On the other hand, we drop the requirement that f be monic, then all of K(R) is integral over R, since a fraction a/b is a root of f(x) = bx a. In fact, we already know some examples of integrally closed domains. Proposition 5.2. UFDs are integrally closed. Proof. Let R be a UFD and let a/b K(R) be integral over R; say (a/b) n + c n 1 (a/b) n c 0 = 0. We assume a and b have no common factor (so a/b is in lowest terms ). We can then multiply the equation above by b n to obtain a n + bc n 1 a n b n c 0 = 0, or a n = bc n 1 a n 1 b n c 0. But the right-hand side is divisible by b, and a and b had no common factor and hence since R is a UFD, a n and b have no common factor. The only way this is possible is if b is a unit. But in that case, a/b is actually in R, so R is integrally closed. On the other hand, we also have a non-example: Proposition 5.3. Z[ 3] is not integrally closed. Proof. Let ω = (1 + 3)/2. Then ω is certainly in the field of fractions of Z[ 3], and it is not in Z[ 3]. Thus if we show ω is integral, Z[ 3] will not be integrally closed. To show this, we can simply note that ω is a root of f(x) = x 2 + x + 1, by the quadratic formula. Remark 5.4. Note that more generally, (1 + d/2 is a root of x 2 + x + (d + 1)/4, so whenever d 3 mod 4 this shows that Z[ d] is not integrally closed. Thus Z[ 3] in some sense lacks unique factorization for a stupid reason that is easily remedied: it s just missing some elements it should have. More precisely, we have the following fact, whose proof is beyond the scope of this class. Theorem 5.5. Let R be a domain. Then the set R of elements of K(R) integral over R is a ring and is integrally closed, called the integral closure of R. That is, if we just take R and throw in all the elements that are integral over it, we can complete R to an integrally closed domain, which now has some hope of being a UFD. Indeed, this works in our 3 example. Theorem 5.6. Let ω = (1 + 3)/2 as above. Then Z[ω] is the integral closure of Z[ 3], and is a UFD. 12

13 Proof. If Z[ω] is a UFD, it is automatically the integral closure of Z[ 3]: indeed, it is integrally closed, and therefore contains the integral closure of Z[ 3]. But ω is integral over Z[ 3], so it is also contained in the integral closure of Z[ 3]. We thus show that Z[ω] is a UFD. We use the same argument as we always have: we show that it is a Euclidean domain under the norm function given by the square of the complex absolute value, by a geometric argument. Now instead of a rectangle we have a parallelogram given by 1 and ω in the complex plane, but the geometric computation still works out. A similar argument shows that Z[(1 + 7)/2] is a UFD and the integral closure of Z[ 7]. More generally, it is always true that for d 3 mod 4 (and squarefree), Z[(1 + d)/2] is the integral closure of Z[ d]. However, the geometric arguments quickly break down as d gets large, and in fact we have the following theorem. Theorem 5.7. Let d > 0 be squarefree. Then the integral closure of Z[ d] is a UFD iff d is 1, 2, 3, 7, 11, 19, 43, 67, or 163. This result is far beyond the scope of this class, and was proven only in Nevertheless, it turns out that there is a generalization of unique factorization that often works in domains that are not UFDs. The idea is to not attempt to factorize elements of the ring, but ideals. This is a natural thing to do because the ideal (a) generated by an element a depends on a only up to multiplication by units (and in fact determines a, up to a unit), and in a PID ideals are exactly the same as elements, up to units. Indeed, this was historically the original reasons that ideals were defined: they are ideal elements that can be substitutes for elements (up to units) for purposes of factorization, but are better-behaved in domains that are not PIDs. To make sense of this, we should first define multiplication of ideals. Definition 5.8. Let I, J R be ideals. Then IJ is the ideal of elements { c n d n : c n I, d n J}. Note that in the case of principal ideals, we have (a)(b) = (ab), since every term of the sum c n d n is divisible by ab, so the whole sum is. More generally, if I is generated by {a i } and J is generated by {b j }, then IJ will be generated by {a i b j }. We also have a definition of a prime ideal, which we have already seen in Exercise 2.6. Definition 5.9. An ideal I R is prime if whenever ab I, then either a I or b I. Note that a principal ideal (p) is prime iff the element p is prime. Let s see an example how using prime ideals instead of irreducible elements might be able to save unique factorization. Example Recall that in Z[ 5], 9 has two different irreducible factorizations, 3 3 and (2+ 5)(2 5). However, these do not give two different factorizations of the ideal (9), since the ideals (3) and (2 ± 5) are not prime! Indeed, since 3 9 but 3 2 ± 5, (3) cannot be prime, and similarly for (2 ± 5. However, it turns out that the non-principal ideals I = (3, 2 + 5) and J = (3, 2 5) are prime. Furthermore, we have IJ = (3, 2 + 5) (3, 2 5) = (9, 6 3 5, , 9). Note that everything on the right-hand side is divisible by 3, so IJ (3). On the other hand, since and are both in IJ, so is their difference 12, so IJ (9, 12) = (3). Thus IJ = (3). A similar computation shows that I 2 = (2 + 5) and J 2 = (2 5). Thus if we think in terms of ideals, the failed uniqueness of factorization 3 3 = (2 + 5)(2 5) becomes just IJ IJ = I 2 J 2, both giving the same factorization of (9) into prime ideals. However, using prime ideals will not always save us. As an example, consider the ring k[t 2, t 3 ], which we earlier saw was not a UFD because t 2 and t 3 are both irreducible but (t 2 ) 3 = (t 3 ) 2. In this case, the only 13

14 prime ideal containing either (t 2 ) or (t 3 ) is the ideal I = (t 2, t 3 ). However, I 2 = (t 4, t 5, t 6 ) = (t 4 ) is already smaller than both (t 2 ) and (t 3 ). This means that (t 2 ) and (t 3 ) cannot even be written as a product of prime ideals at all, let alone uniquely. It turns out the problem is that t 3 /t 2 = t should be an element of this ring, so that instead of I = (t 2, t 3 ) we could have I = (t), and then everything would work. More precisely, this ring is not integrally closed, since t is a root of f(x) = x 2 t 2 and hence integral. It turns out that being integrally closed is almost all you need to be able to use ideals to get unique factorization. Theorem Let R be an integrally closed domain which is Noetherian and in which every prime ideal except {0} is maximal (a ring satisfying these hypotheses is called a Dedekind domain. Then every ideal I R factors uniquely as a product P di i where the P i are prime ideals. Again, it would take more time (or more chilies) than we have to prove this theorem. Here Noetherian is a technical condition that says that R is not too large and pathological (more precisely, a ring is Noetherian if every ideal can be generated by a finite set of elements). The other condition, that every prime ideal is maximal, is satisfied by any PID and by the integral closure of Z[ d] for any d, and it turns out that this condition together with being a UFD actually implies that your ring is a PID. However, we have already seen that there are UFDs that are not PIDs and hence do not satisfy this property, namely polynomial rings over UFDs that are not fields. 5.1 Exercises For the exercises, you need the following result, which you can try to prove if you re interested (though it is hard without using tools we have not introduced in this class). Theorem Let R S be domains a S be integral over ar, and let f(x) R[x] be of minimal degree such that f(a) = 0. Then the leading coefficient of f is a unit in R. (The definition of integral we gave before only stipulates that there is some monic polynomial f such that f(a) = 0.) Exercise 5.1. Show that if d is square-free and d 3 mod 4, then Z[(1+ d)/2] is integrally closed. (Hint: The field of fractions is Q[ d]. Given an element z = a + b d with a, b Q, find a quadratic polynomial that z is a root of, and show this polynomial can be made a monic polynomial in Z[x] iff z Z[(1+ d)/2]. Conclude that Z[(1 + d)/2] is the integral closure of Z in Q[ d], and hence is integrally closed by Theorem 5.5.) Exercise 5.2. Show that if d is square-free and d 1 mod 4, then Z[ d] is integrally closed. (Hint: This is similar to the previous exercise.) Exercise 5.3. Show that Z[(1 + 7)/2] is a Euclidean domain, and hence a PID and UFD. 14

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