Math 210B: Algebra, Homework 1

Math 210B: Algebra, Homework 1 Ian Coley January 15, 201 Problem 1. Show that over any field there exist infinitely many non-associate irreducible polynomials. Recall that by Homework 9, Exercise 8 of last quarter, for any nonzero commutative ring R, R[X] has infinitely many prime ideals. Hence the same is true for a field F and F [X]. Further recall that F [X] is a PID. Therefore to each prime ideal in F [X] we may assign a generator p(x). By definition, this generator is prime, hence irreducible. Further, for two distinct prime ideals p and q, their generators p(x) and q(x) are non-associate since p q. Hence F [X] has infinitely many non-associate irreducible polynomials. Problem 2. Show that Z[ 2] is a UFD. We will show that Z[ 2] is, in fact, a Euclidean domain. First, let N : Z[ 2] \ {0} N so that a + b 2 a 2 2b 2. We claim this is the requisite Euclidean function. The required property is: For all α, β Z[ 2], β 0, γ, δ Z[ 2] such that α = βγ + δ and δ = 0 or N(δ) < N(β). So let α = a 0 + a 1 2, β = b0 + b 1 2. Then viewing Z[ 2] Q( 2) as a subring of its quotient field, let α β = c 0 + c 1 2, ci Q. Then choose p, q Z closest to c 0, c 1 so that p c 0 1/2 and q c 1 1/2. Then let γ = p + q 2. Let η = (c 0 p) + (c 1 q) 2 so that α/β = γ + η. Then multiplying through, α = γβ + ηβ. Since α Z[ 2] and γβ Z[ 2], we must have ηβ Z[ 2]. Hence let δ = ηβ. Then we need to show N(δ) < N(β). Examining η, we see N(η) = (p c 0 ) 2 2(q c 1 ) 2 (p c 0 ) 2 + 2(q c 1 ) 2. 1

Since p c 0, q c 1 1/2, we have N(η) (1/2) 2 + 2(1/2) 2 = 3/. Therefore since N(δ) = N(ηβ), in particular N(δ) 3 N(β) < N(β). We use here the additional fact that the norm is multiplicative, noting ) ( N(αβ) = N ((a 0 + a 1 2)(b0 + b 1 2) = N (a 0 b 0 + 2a 1 b 1 ) + (a 0 b 1 + a 1 b 0 ) ) 2 = (a 0 b 0 + 2a 1 b 1 ) 2 2(a 0 b 1 + a 1 b 0 ) 2 = (a 2 0 2b 2 0)(a 1 2b 2 1) = N(α)N(β). Therefore Z[ 2] is a Euclidean domain, and a fortiori it is a UFD. Problem 3. Show that Z[ 5] is not a UFD. Since Z[ 5] C, the multiplicative complex norm makes sense in this ring. Note that for a + b 5 Z[ 5], we have a + b 5 = a 2 + 5b 2. Now, we claim that 6 = 2 3 = (1 + 5)(1 5) are two factorisations of 6 into irreducible elements. Note that an element α Z[ 5] is reducible if and only if α is a product of two integers > 1 of the form a 2 + 5b 2 for a, b Z. In our case, we have 2 =, whose only nontrivial factorisation is 2 2. Since 2 a 2 +5b 2 for any integral a, b Z, 2 is irreducible. Similarly the only nontrivial factorisation of 3 = 9 is 3 3, and 3 a b +5b 2 for integral a, b Z. Finally, we have 1+ 5 = 1 5 = 6 has the only nontrivial factorisation 6 = 2 3, and we have shown that no elements of Z[ 5] has norm 2 or 3. Hence all these numbers are irreducible. Since some element of Z[ 5] admits two factorisations into irreducibles, the ring is not a UFD. Problem. Let S R be a multiplicative subset. Prove that the natural ring homomorphism R S 1 R is an isomorphism if and only if S R. Let ϕ be the natural ring homomorphism, r rs/s for a fixed s S. First, assume that R is zero ring. Then S 1 R is also the zero ring (since we must have S = ), so ϕ is an isomorphism. Henceforth let R be a ring with 1 0. Then we cannot have 0 S, else S 1 R is the zero ring, so ϕ could not be an isomorphism since R 0. Therefore assume that 0 / S hereafter. We claim that ϕ is injective if and only if S contains no zero divisors. Suppose that ϕ is injective and that a S is a zero divisor with ab = 0, b 0. Note that ra/a = rs/s since ras rsa = 0. Therefore ϕ(b) = bs s = ba a = 0 a = 0 S 1 R 2

which is a contradiction. Therefore S contains no zero divisors. Conversely, suppose that S contains no zero divisors and ϕ(r) = 0. Then rs s = 0 s t S such that t(rs2 ) = 0. Since t is not zero and not a zero divisor, we must have rs 2 = 0. Since s is not a zero divisor, we must have r = 0. Therefore ϕ is injective. Now we claim that ϕ is surjective if and only if S R. Suppose that ϕ is surjective and suppose that t S \ R. Then since ts/s s/ts = 1 S 1 R, we must have 1 R = ϕ 1 (ts/s s/ts) = ϕ 1 (ts/s)ϕ 1 (s/ts) = t τ for some τ R. But since t / R, this is a contradiction. Therefore S R. Conversely, assume that fact. Then let r/t S 1 R. Then since t R, ϕ(rt 1 ) = rt 1 s s = r t because rt 1 st rs = 0. Therefore every element of S 1 R has a nonempty preimage, so ϕ is surjective. Putting these together, suppose ϕ is an isomorphism. Then since ϕ is surjective, S R. Conversely, suppose that S R. Then ϕ is surjective. Further, since units and zero divisors are mutually exclusive, S contains no zero divisors, so ϕ is injective. Hence ϕ is an isomorphism. This completes the proof. Problem 5. Show that the ring of Laurent polynomials a n X n +... + a m X m with integral coefficients is a localisation of Z[X]. Let S = {X n : n 1}. Let S 1 Z[X] = Z[X] X be the localisation at this multiplicatively closed set. Then we claim Z[X] X = Z[X, X 1 ], the ring of Laurent polynomials. Note that any polynomial f = n i= m a ix i we may write n+m f = X m a i m X i. Let i : Z[X] Z[X, X 1 ] be the inclusion map, and note that X m Z[X, X 1 ] for all m 1. Therefore i(s) Z[X, X 1 ], so by the universal property, i extends to S 1 Z[X] = Z[X] X. Let ϕ be this map. It is still injective, and we see it s surjective since for any f = n i= m a ix i, ( n+m ϕ a ) i mx i n+m = X m a X m i m X i = f. So every polynomial has a nontrivial preimage. Therefore ϕ is a bijective ring homomorphism, so Z[X, X 1 ] = Z[X] X. 3

Problem 6. Prove that if R is a UFD and 0 / S R is a multiplicative subset, then S 1 R is also a UFD. First note that if 0 S, then S 1 R = 0 is identically the zero ring, so it is not a domain, hence not a UFD. Now assume 0 / S. Let r/s S 1 R. Then if r S, we have r/s s/r = 1, so r/s (S 1 R). Therefore r/s is its own unique factorisation as the product of itself and an empty product of primes. Now suppose r R \ S. Then we can write r s = 1 s r 1. Since r admits a unique factorisation in R, say r = p 1 p n, this factorisation injects into S 1 R. Problem 7. Let R = Z/6Z. Prove that S = { 1, 2, } is a multiplicative subset in R and S 1 R = Z/3Z. We first prove that S is closed under multiplication. Clearly multiplication by 1 preserves S. Further, 2 2 = S, 2 = 2 S, and = 2 S, so S is a multiplicative set. We know examine the structure of S 1 R. Write this ring F. If F has three elements (i.e. three equivalence classes of elements), then by a previous exercise, S 1 R = Z/3Z since there is only one ring with three elements up to isomorphism. The elements are (suppressing the bar notation) { 0, 1, 1 2, 1, 2 1, 3 1, 3 2, 3, 1, 5 1, 5 2, 5 } where we know 0/n = 0 F and n/n = 1 F, and we reduce 2, in the usual way. We will show 2 all of these elements are equivalent to 2 = 2/1, 1 = 1/1, or 0 = 0/1. Checking some against 2, and remembering 6 0 6, 2 (1 1 2 2) = 0, 2 (2 1 5 1) = 0, 2 (2 5 1) = 0 For some of the others, we check against 1: 2 (1 1 1) = 0, 2 (1 1 1) = 0, 2 (1 5 2 1) = 0 And the remainder we check against 0: 2 (3 1 0 1) = 0, 2 (3 1 0 2) = 0, 2 (3 1 + 0 ) = 0 This shows F has only elements, so we must have S 1 R = F = Z/3Z. Problem 8. Let R be an integral domain with quotient field F. Show that if T is an integral domain such that R T F, then the quotient field of T is isomorphic to F.

We will show that any fraction r/s, r, s T, is equivalent to a fraction a/b with a, b R. Since T F, we can express r = α/β and s = γ/δ with α, β, γ, δ R. Then α r s = β γ δ = αδ βγ because ( α β βγ γ ) δ αδ = (αβγ αβγ) = 0. Therefore since αδ, βγ R, r/s F. Since this holds for every fraction r/s, the quotient field of T is contained in F. But since R T, the quotient field of T must contain F. Since we have shown double inclusion, R and T have the same quotient field. Problem 9. Let f : R S be a homomorphism of commutative rings. Let p be a prime ideal of S. Prove that f 1 (p) is a prime ideal of R. Give an example of a maximal ideal m of S such that f 1 (m) is not maximal. Let f 1 (p) = {a R : f(a) p} = q. It suffices to show that if ab q, then a q or b q. For ab q, we have f(ab) = f(a)f(b) p. Since p is prime, then f(a) or f(b) is in p. Hence f 1 (f(a)) = a or f 1 (f(b)) is in q. Now suppose R = Z and S = Q, where f : Z Q is the inclusion map. Then 0 Q is a maximal ideal,but f 1 (0) = 0 is not a maximal ideal in Z because Z is not a field. Problem 10. Show that if the ring R = Z[X 1, X 2,...] is a UFD. Is R is a noetherian ring? Suppose f R. Then only finitely many X i appear in f, and without loss of generality let these be X 1,..., X n. Then we claim that f is irreducible in R if and only if f is irreducible in R n = Z[X 1,..., X n ]. The forward direction is clear. For the converse, suppose that f is irreducible in R n but reducible in R. Then let f = g h, and without loss of generality assume g R \ R n. Let X N be a variable with N > n, and let b N be the coefficient of some term with X N appearing in g. UGH R is not noetherian. Consider the ideal a i = (X 1,..., X i ), generated by the first i variables. Then the chain a 1 a 2... does not stabilise. Since R does not satisfy the ascending chain condition, it is not noetherian. 5