# Example: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x)

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1 Math 4010/5530 Factorization Theory January 2016 Let R be an integral domain. Recall that s, t R are called associates if they differ by a unit (i.e. there is some c R such that s = ct). Let R be a commutative ring (with 1) and a, b R. a divides b iff (b) (a). Why? f a divides b, then ak = b for some k R. So br = akr. Thus multiples of b are multiples of a. This means (b) (a). For the converse notice that (b) (a) means b (b) (a) so that b (a) and so b = ak for some k R. This means (a) = (b) iff a divides b and b divides a. n an integral domain (a) = (b) iff a and b are associates. Why? f a and b are associates, a = bu for some unit u (so b divides a) and thus also b = au 1 (so a also divides b). This then means (a) = (b). Conversely, suppose (a) = (b). Then a (b) and b (a). Therefore, a = bk and b = al for some k, l R. This means a = bk = alk. Now if a = 0, then b = 0 and so a = b = 0 are associates. f a 0, then since R is an integral domain, we can cancel a and get 1 = lk. Thus a = bk where k is a unit. This means a and b are associates. A non-zero, non-unit element r R is called irreducible if given r = ab for some a, b R then we have that either a or b is a unit (making the other element an associate of r). n other words, r has no interesting factorizations. A non-zero, non-unit element p R is called prime if whenever p divides ab (for some a, b R) then either p divides a or p divides b. This definition guarantees that primes satisfy the conclusion of Euclid s lemma. Let R be a commutative ring (with 1). Let P R and P R. We say that P is a prime ideal if ab P implies that either a P or b P. Directly from the definition: p 0, (p) is a prime ideal iff p is a prime element. Also, (0) is a prime ideal iff our ring is an integral domain. Theorem: Let R be a commutative ring (with 1) and P R. P is a prime iff R P is an integral domain. proof: Suppose P is a prime ideal. Since P is a proper ideal, R/P is a non-trivial commutative ring (with 1). Suppose a + P, b + P R/P and (a + P )(b + P ) = 0 + P. Then ab + P = P so ab P. Since P is prime, either a P or b P so that either a + P = 0 + P or b + P = 0 + P. This shows that R/P has no zero divisors. Therefore, R/P is an integral domain. Conversely suppose R/P is an integral domain. Since R/P is a non-trivial ring, P must be a proper ideal. Suppose ab P. This implies that (a + P )(b + P ) = ab + P = 0 + P. Since R/P is an integral domain, it has no zero divisors and thus either a + P = 0 + P or b + P = 0 + P. Thereofore, either a P or b P and so P is prime. Example: This theorem is the easiest way to test an ideal (or an element) is prime. Z[x] (x) = Z is an integral domain. Therefore, (x) is a prime ideal in Z[x]. This means x is a prime element of Z[x] as well. Let R be a commutative ring (with 1). Let M R and M R. We say that M is a maximal ideal if given R and M R, then either M = or = R. Theorem: Let R be a commutative ring (with 1) and M R. M is maximal iff R M is a field. proof: Let M be maximal. Then M is a proper ideal so R/M is a non-trivial commutative ring (with 1). Let r + M be a non-zero element of R/M. Consider = (r) + M = {rs + m s R and m M}. is an ideal of R with M. Also, r but r M (otherwise r + M = 0 + M), so we have M. But M is a maximal ideal, so = R. Therefore, 1 so that 1 = rs + m for some s R and m M. So 1 + M = (rs + m) + M = rs + M = (r + M)(s + M). Thus r + M is a unit. This means R/M is a field. Conversely, suppose R/M is a field. Since fields are non-trivial rings, M must be a proper ideal. Suppose is an ideal of R such that M R. Moreover, assume M. This means there is some r such that r M. Therefore r + M 0 + M. But R/M is a field so r + M is a unit, say (r + M)(s + M) = 1 + M. Since 1 + M = rs + M, 1 = rs + m for some m M. Now r so rs ( is an ideal) also m M so m (since M ). Therefore, 1 = rs + m. and so = R. This means M is a maximal ideal. 1

2 Corollary: For ideals of commutative rings, maximal implies prime. Example: This (like the theorem about prime ideals) is the easiest way to test to see if an ideal is maximal. Notice that since Z[x] (x) = Z is not a field, (x) is not a maximal ideal of Z[x]. n fact, (x) (x, 5) = {f(x)x + g(x)5 f(x), g(x) Z[x]}. Notice, however, that Z[x] (x, 5) = Z (5) = Z 5 is a field. This means that (x, 5) is a maximal (and thus also prime) ideal of Z[x] (and (5) is a maximal/prime ideal of Z). The relationship between prime and irreducible? Lemma: n an integral domain, prime implies irreducible. proof: Let p be a prime in an integral domain R. By definition p is non-zero and not a unit. Let p = ab for some a, b R. Then since p divides ab we have that it either divides a or b. WLOG assume p divides a. This means there is some k R such that pk = a and so p = ab = pkb. But R is an integral domain and p 0 so we can cancel and get 1 = kb (i.e. b is a unit). This means p is irreducible The converse of the above statement is not always true. The implication irreducible implies prime is almost equivalent to requiring uniqueness of factorizations. Example: Consider Z[ 5]. Define N(a + b 5) = (a + b 5)(a b 5) = a 2 + 5b 2 (a norm for this ring). t can be shown that N(xy) = N(x)N(y) and that if u is a unit, then N(u) = 1. Notice that (1 + 5)(1 5) = 6 = 2 3 (a non-unique factorization of 6). However, N(1 + 5) = 6 which does not divide N(2) = 4 or N(3) = 9. This means that even though divides 6 it does not divide 2 or 3, so it is not a prime element. On the other hand, if = xy then 6 = N(xy) = N(x)N(y). Since a 2 + 5b 2 2 or 3 for any choice of integers a and b, there are no elements of norm 2 or 3. Thus N(x) = 1 or N(y) = 1. This means either x or y is a unit. Since N(1 + 5) = 6 1, isn t a unit. Thus it s irreducible. So is irreducible but not prime (in Z[ 5]). An integral domain R is called a unique factorization domain (UFD) if every non-zero, non-unit element can be factored uniquely into irreducible elements. Specifically, given r R where r 0 and r R (a non-zero, non-unit), there exists irreducible elements a 1,..., a n R such that r = a 1 a n. Moreover, if r = a 1 a n and r = b 1 b m where a 1,..., a n, b 1,..., b m are irreducible, then n = m and there is some permutation σ S n such that a i is an associate of b σ(i) for i = 1,..., n. n other words, factorization into irreducibles is unique up to order and associates. Example: Z is a nice example of a UFD. Notice that 6 = 2 3, so 6 factors into irreducibles (= primes in this ring). But also 6 = ( 3) ( 2). This second factorization isn t really all that different. t s just the first factorization reordered and our primes have been swapped out with associates. A similar example would be x 2 = x x = 2x x/2 in R[x] (we typically favor the first factorization). Lemma: n a UFD, prime = irreducible. proof: n an integral domain, prime always implies irreducible. Suppose that p is irreducible. Then p is non-zero and not a unit. Suppose p divides ab so that pk = ab for some k. f ab = 0 then either a = 0 or b = 0 and so p divides either a or b. Now assume, a, b 0 (and so k 0). f a is a unit, then pa 1 = b and so p divides b (likewise for b a unit). Suppose that neither a nor b is a unit. Then a and b can be factored into irreducibles, say a = a 1... a n and b = b 1 b m. This means that pk = ab = a 1... a n b 1... b m is a factorization into irreducibles. Recall that p is irreducible so it must be an associate of one of the irreducible factors. WLOG assume p is an associate of a n. Then because a n divides a so does p. Thus p is prime. Lemma: n the definition of UFD, uniqueness of factorizations can be replaced by prime = irreducible. proof: We just showed that prime = irreducible in a UFD. Conversely, suppose R is an integral domain with factorizations and that all irreducibles are prime. Let r = a 1... a n = b 1... b m for irreducibles a 1,..., a n, b 1,..., b m. Then a n divides r = b 1... b m and since a n is irreducible, we have assumed it s prime. Thus a n divides b j for some j = 1,..., m. WLOG assume a n divides b m (reorder the b s if necessary). But these are irreducibles. Thus if one divides another, they re associates. Now cancel both off and continue by induction. Uniqueness now follows. mportant Theorem: R is a UFD iff R[x] is a UFD. [The proof of this theorem is straight forward relying heavily on a result known as Gauss lemma, but is long enough think we will skip it unless there is enough interest to get into the details.] Example: Since Z is a UFD, so is Z[x]. This then means (Z[x])[y] = Z[x, y] is a UFD as well. 2

3 Let R be a ring. R has the ascending chain condition (ACC) on ideals if given ideals, 1, 2, 3, R such that 1 2 3, we have that there is some N > 0 such that N = N+1 = (every increasing chain of ideals eventually stops growing that is it eventually stabilizes). f every increasing chain of principal ideals eventually stabilizes, the ring has the ascending chain condition on principal ideals (ACCP). Lemma: n an integral domain, the ACCP implies factorizations exist. proof: Suppose there is a non-zero, non-unit element which has no factorization, call it x 1. Since x 1 is not irreducible (otherwise, it is its own factorization contradiction), there must exist non-unit elements a 1, b 1 such that x 1 = a 1 b 1. Now if both a 1 and b 1 have factorizations, then so does x 1 = a 1 b 1 (multiply the factorizations together). This means either a 1 or b 1 fails to have a factorization, call the offending element x 2. Continuing in this fashion we get x 1, x 2,... where x n+1 is a divisor of x n but not an associate (recall that for each n, x n = a n b n where neither a n nor b n is a unit and x n+1 is either a n or b n ). This means that (x 1 ) (x 2 ) (x 3 ) (since each generator is a divisor of the previous one) and (x n ) (x n+1 ) for each n = 1, 2,... (since generators are not associates). Thus we have created a infinitely ascending chain of principal ideals this is a contradiction! Therefore, every element must have a factorization. Example: t turns out that Z[ 5] has the ACCP (and even the ACC). However, we already saw a non-prime, irreducible element so it is not a UFD. This ring has factorizations, just not unique ones. Theorem: Let R be an integral domain. R is a UFD iff prime = irreducible in R and R has the ACCP. proof: Suppose that R has the ACCP. Then R has factorizations. Then if factorizations exist and prime = irreducible we already know R is a UFD. Conversely, suppose R is a UFD. We already know that in a UFD prime = irreducible. Next, suppose (0) (x 1 ) (x 2 ) is a chain of principal ideals and that it never stabilizes. WLOG assume each containment is proper. Notice that for each n 1, x n+1 must divide x n, but (x n ) (x n+1 ) so x n and x n+1 are not associates. Next, factor x n+1 (it isn t zero and it cannot be a unit otherwise (x n+1 ) = R and so our chain stabilizes after n + 1), say x n+1 = a 1... a k. Since this divides x n, we have x n+1 y = x n for some y. We know that y is not a unit (otherwise x n and x n+1 are associates), so we can factor y, say y = b 1... b l. Thus x n = x n+1 y = a 1... a k b 1... b l. Therefore, x n has k + l irreducible factors whereas x n+1 only has k irreducible factors. Long story short, for each n 1, x n+1 has fewer factors than x n. This is impossible since x 1 has finitely many factors the ideals in our chain can only grow finitely many times! Therefore, R has the ACCP. t is interesting to note that not all UFDs have the ACC (the ascending chain condition on all ideals). Example: f R is a UFD, then so is R[t 1, t 2,... ] (polynomials in t 1, t 2,... and coefficients in R). So the ring of polynomials in countably many variables with coefficients in R has the ACCP. However, the full ACC fails. We can see this by considering the infinitely ascending chain: (t 1 ) (t 1, t 2 ) (t 1, t 2, t 3 ). An integral domain R is called a principal ideal domain (PD) if every ideal is principal. This means that if R, then there exists some a R such that = (a) = {ra r R}. Theorem: PD implies UFD. proof: Let R be a PD. Suppose p R is irreducible. Since p is not a unit, we have (p) R. Next, suppose that (p) R and (p). Since R is a PD, = (a) for some a R. So p (p) (a). Thus there is some y R such that p = ay. But p is irreducible, thus either a or y is a unit. f y is a unit, p and a are associates and so (p) = (a) = (contradiction). Thus a is a unit and so = (a) = R. This shows that (p) is a maximal ideal. Therefore, (p) is a prime ideal. p 0 (since it s irreducible) and so p is prime. We have now shown that prime = irreducible in a PD. Next, consider a chain of (principal) ideals: (x 1 ) (x 2 ). Let = (x n ). t is straightforward to verify that R. But R is PD so = (a) for some a R. Consider n=1 a = (x n ) so a (x k ) for some k. This means = (a) (x k ) (x k+1 ). Therefore, the chain n=1 stabilizes at k. Thus R has the ACCP (= ACC since all ideals are principal). Therefore, R is a UFD. Example: Both Z and F[x] (where F is a field) are PDs. This then means that they must also be UFDs. An integral domain R equipped with a norm δ : R {0} Z 0 is called a Euclidean domain if... (1) For each a, b R such that b 0 there exists q, r R such that a = bq + r and either r = 0 or δ(r) < δ(b). (2) For each a, b R {0}, δ(a) δ(ab). Note: Some authors omit condition (2). n the end it is more-or-less redundant. f an integral domain posses a norm N that only satisfies (1) and one defines δ(a) = min{n(ax) x R {0}} for each a R {0}, then it can be shown that δ satisfies both (1) and (2). We choose to work with nice norms. 3

4 A lot can be done in Euclidean domains. wrote a paper with Tyler Bradley (a former ASU student) which proves the existence of partial fraction decompositions in the context of Euclidean domains. Theorem: Every Euclidean domain is a PD. proof: Let R. f = {0} then = (0) is principal. Now suppose {0} and consider the set of all norms of elements of : S = {δ(b) b {0}}. Since {0}, S is non-empty. By the well ordering principle, S (a non-empty subset of non-negative integers) has a least element: l = δ(b) for some b {0}. Obviously since b, (b). Suppose a. We are in a Euclidean domain so there exists q, r R such that a = bq + r and r = 0 or δ(r) < δ(b). Notice that r = a bq (since a, b ). So if r 0, δ(r) is smaller than δ(b) and r {0} (contradiction). Thus r = 0 and so a = bq (b). Therefore, (b) and so = (b). We have shown that every ideal is principal. We know have that every Euclidean domain is a PD and thus is also a UFD. We already know examples of UFDs which aren t PDs (like Z[x] or R[x, y]). t is natural to ask if there are any PDs which aren t Euclidean domains? The answer is, Yes, but proving this is the case is a bit more difficult. Since for the most part we tend to work with PDs that are also Euclidean domains, assuming a ring has a division algorithm doesn t take away that much. Many textbooks prove propositions in the context of Euclidean domains when (with a slight adjustment) their proofs will work for more general PDs. A great example of this is the classification of finitely generated modules over a PD. Michael Artin s Algebra proves this classification theorem for Euclidean domains when he could just as well have done it for PDs. The real advantage of working with Euclidean domains is that you then get more constructive results (and proofs tend to look more algorithmic). Example: Z[η] where η = is a PD but not a Euclidean domain. A good reference for this fact is 2 Dummit and Foote s Abstract Algebra (2 nd edition). [This is established on page 277 just after Proposition 5.] Example: Z, F[x] (for any field F), and Z[i] (the Gaussian integers) are all examples of Euclidean domains (and thus are PDs and UFDs as well). Just for fun. When are quotients and remainder unique? Theorem: Let R be a Euclidean domain equpped with norm δ. (1) Quotients and remainders are unique in R iff δ(a + b) max{δ(a), δ(b)} for all a, b R {0} such that a + b 0. (2) f quotients and remainders are unique, then the units of R along with zero form a field K. Moreover, either R = K or R = K[x]. Essentially, this says that quotients and remainders are unique only for polynomials with field coefficients. Example: thought quotients and remainders where unique for Z! Depends. The norm for Z is δ(n) = n. This does not satisfy the condition of the above theorem. Also, notice that 5 = (2)2 + 1 (here q = 2 and r = 1 where δ(r) = 1 < δ(2) = 2) but also 5 = (3)2+( 1) (here q = 3 and r = 1 where δ(r) = 1 = 1 < δ(2) = 2). We usually think of quotients and remainders being unique for integer division because we demand that our remainders are always non-negative being positive or negative isn t part of the Euclidean domain structure of Z! n other words, quotients and remainder are not unique for Z, the Euclidean domain, but they are unique for the ordered ring Z when we demand non-negative remainders. Let R be an integral domain and suppose a, b R where ab 0 (they re not both zero). d R is a greatest common divisor (GCD) of a and b iff (1) d divides both a and b (it s a common divisor) and (2) given any common divisor c (c divides both a and b), we have that c also divides d. An GCD of a and b is commonly denoted gcd(a, b) or (a, b) Likewise, we say that l R is a least common multiple (LCM) of both a and b iff (1) a and b both divide l (it s a common multiple) and (2) given any common multiple c (a and b both divide c), we have that l divides c. An LCM of a and b is commonly denoted lcm(a, b) or [a, b]. Lemma: GCDs and LCMs exist in UFDs. proof: Let a = r s1 1 rsn n and b = r t1 1 rtn n be factorizations of a and b into distinct (meaning non-associate) prime (aka irreducible) powers (we allow exponents to be zero). Then it s straight-forward to show that a GCD is r min{s1,t1} 1 rn min{sn,tn} and an LCM is r max{s1,t1} 1 rn max{sn,tn}. Notice from the above proof we get that the product of a GCD with a LCM is an associate of ab. n general, GCDs and LCMs aren t unique. They are only determined up to associates unless we work in Z where we 4

5 demand them to be positive or F[x] (polynomials with field coefficients) where we demand them to be monic (have leading coefficient 1). Lemma: Let R be a PD. Let (a) + (b) = (d) and (a) (b) = (l). Then d = gcd(a, b) and l = lcm(a, b). So in a PD we have that a GCD can be expressed as a linear combination of a and b: d = ax + by for some x, y. This is not true in a general UFD. t is then interesting to ask how we can find such x, y to get gcd(a, b) = ax + by. The answer is the Extended Euclidean Algorithm. Lemma: Let R be an integral domain, a, b, q, r R and a = bq + r. Then the common divisors of a and b are exactly the same as the common divisors of b and r. n particular, a GCD of a and b is also a GCD of b and r (and vice-versa). proof: Let c divide both a and b. Then c divides r = a bq. Conversely, let c divide both b and r. Then c divides a = bq + r. Since the pairs a, b and b, r share the same common divisors, they must have the same greatest common divisors. Euclidean Algorithm: Let R be a Euclidean domain. Let a, b R with b 0. Then compute: a = bq 1 + r 1 where r 1 = 0 or δ(r 1 ) < δ(b). Next, compute b = r 1 q 2 + r 2 where r 2 = 0 or δ(r 2 ) < δ(r 1 ). n general, compute: r n = r n+1 q n+2 + r n+2 where r n+2 = 0 or δ(r n+2 ) < δ(r n+1 ). Suppose r N+1 = 0 where r N 0. Then gcd(a, b) = r N. Moreover, running these calculations backwards yields x, y R such that ax + by = r N. proof: First, note that this process of repeated division must eventually stop since δ(b) > δ(r 1 ) > δ(r 2 ) > (it can go on no more than δ(b) steps). Next, notice that a, b and b, r 1 must have the same GCDs (according to the above lemma). Likewise, b, r 1 and r 1, r 2 have the same GCDs. Continuing in this fashion we have that a, b and r N, r N+1 = 0 share the same GCDs. However, the common divisors of r N and 0 are just the divisors of r N (everything divides 0 since x0 = 0). Thus r N is a GCD of r N and 0 (and so is a GCD for a and b). To compute x, y such that ax + by = r N we need to use our divisions last to first successively substituting back until we get our desired result. We have r N 2 = r N 1 q N + r N so r N = (1)r N 2 + ( q N )r N 1. Next, r N 3 = r N 2 q N 1 + r N 1 so r N 1 = (1)r N 3 + ( q N 1 )r N 2. We can now plug this expression in for r N1 resulting in r N = xr N 2 + yr N 3. The next step will rewrite r N 2 in terms of r N 3 and r N 4. After cycling back through all of our divisions (finitely many steps), we will end up with r N = ax + by for some x, y. This also yields an effective way of computing LCMs. Since a GCD times an LCM is an associate of ab, we get that ab divided by a GCD is an LCM. Example: a = 1 + 5i, b = 3 + i Z[i]. Then a = bq 1 + r 1 where q 1 = 1 i and r 1 = 1 + i (notice that δ(r 1 ) = = 2 < δ( 3 + i) = ( 3) = 10. Next, (1 + i)( 1 + 2i) = 3 + i so b = r 1 q 2 + r 2 where r 2 = 0 and q 2 = 1 + 2i. Therefore, 1 + i is a GCD of 1 + 5i and 3 + i. Also, 1 + i = ax + by where x = 1 and y = 1 + i. Example 2: x 3 2x 2 + x 2 divided by x 2 4 is x 2 remainder 5x 10. Next, x 2 4 divided by 5x 10 leaves a remainder of 0. Therefore, gcd(x 3 2x 2 + x 2, x 2 4) = 5x 10 or x 2. Since x 3 2x 2 + x 2 and x 2 4 aren t relatively prime, it turns out that x is a zero divisor in Q[x] where = (x 3 2x 2 + x 2). Let us show this explicitly. We know the gcd is x 2. Divide x 3 2x 2 + x 2 by x 2 and get x Notice that neither x 2 4+ nor x are 0+. However, (x 2 4+)(x 2 +1+) = 0+ since x 2 4 is divisible by x 2 and (x 2)(x 2 + 1). Example 3: Let = (x 3 2). Consider x 2 + 3x Q[x]. Let s run the Euclidean algorithm on a(x) = x 2 + 3x + 2 and b(x) = x 3 2. Dividing we get (x 3 2) = (x 2 + 3x + 2)(x 3) + (7x + 4). Next, dividing x 2 + 3x + 2 by 7x + 4 we get (x 2 + 3x + 2) = (7x + 4)(x/7 + 17/49) + (30/49). The next division must yield a zero remainder. Thus x 2 + 3x is a unit in Q[x]/. Now run it backwards, 30/49 = (x 2 + 3x + 2) (x/7 + /17/49)(7x + 4) = (x 2 + 3x + 2) (x/7 + 17/49)[(x 3 2) (x 3)(x 2 + 3x + 2)]. This means that 30/49 = ( x/7 17/49)(x 3 2) + (x 2 /7 4/49x 2/49)(x 2 + 3x + 2). Multiplying both sides by 49/30 yields 1 = ( 7 30 x ) (x 3 2) + ( 7 30 x x 1 15 ) (x 2 + 3x + 2) This means that (x 2 + 3x ) 1 = 7 30 x x in Q[x]. Here s an interesting interpretation of this result... Let x = 2 1/3 (a root of x 3 2). Then this calculation tells us that 1 2 2/ /3 + 2 = / /

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