## M953 Homework 2

Due Friday, February 8, 2002

[1] Let S be the ring C[x,y] of complex polynomials in two variables, and let R be the ring R[x,y] of real polynomials in two variables. Let f = y + xy - x2 and g = y2 + xy2 - x3, let J be the ideal (f,g) in S, and let I be the ideal (f,g) in R.
• (a)
• (i) Determine V(J) in A(C2).
• (ii) Conclude that Rad(J) = (x,y), where (x,y) means the ideal in S generated by x and y.
• (b)
• (i) Explicitly find an n and m such that xm and yn are in I.
• (ii) Conclude that Rad(I) = (x,y), where now (x,y) means the ideal in R generated by x and y.
Solution:
• (a)
• (i) Claim: V(J) = {(0,0)}.
Since f(0,0)=0=g(0,0), we see {(0,0)} is in V(J). On the other hand, if g - yf = (y - x)x2 is zero, then either y = x or x = 0. If x = 0, then y = x implies y = 0. If y = x, then 0 = f = y + xy - x2 = x + xx - x2 = x, so x = 0 and hence y = 0, as before. Thus if p is in V(J) then p = (0,0).
• (ii) Rad(J) = I(V(J)) by the Nullstellensatz, and V(J) = {(0,0)} so I(V(J)) = I({(0,0)}). But I({(0,0)}) contains (x,y) and (x,y) is maximal, so (since I({(0,0)}) can't be the whole polynomial ring since 1 is clearly not in I({(0,0)})) we see I({(0,0)}) = (x,y).
• (b)
• (i) By playing around here I found: -(g - yf) - x(g -(y + x)f) = x3 and y(g - (y + x)f) + g - (g - yf) - x(g -(y + x)f) = y2. Thus x3 and y2 are in I.
• (ii) By (i) we see that (x,y) is in Rad(I). But (x,y) is maximal, so either (x,y) = Rad(I) or R[x,y] = Rad(I), and clearly 1 is not in Rad(I) (else 1 would be in I), so we must have (x,y) = Rad(I).
[2] If k is an algebraically closed field, use the Weak Nullstellensatz to show that every maximal ideal of k[x1, ... ,xn] is of the form (x1 - a1, ... , xn - an), for some elements ai of k.

Solution: Let J be a maximal ideal. Then the Weak Nullstellensatz says that V(J) is not empty, so let a = (a1, ... , an) be a point of V(J). Thus J is contained in I({a}). Now, I({a}) contains (x1 - a1, ... , xn - an) but is not k[x1, ... ,xn], so (since in class we saw that (x1 - a1, ... , xn - an) is maximal) we must have I({a})= (x1 - a1, ... , xn - an). Thus J is contained in the maximal ideal (x1 - a1, ... , xn - an), so J must equal it.

[3] The Weak Nullstellensatz says that when k is an algebraically closed field, the following statement holds: "If J is a proper ideal of k[x1, ... ,xn], then V(J) is nonempty." When n = 1, show that the quoted statement is equivalent to saying "Every nonconstant polynomial in k[x1] has a root." [You may assume that k[x1] is a PID.]

Solution: If f is a nonconstant polynomial in k[x], then J = (f) is a proper ideal so by the Weak Nullstellensatz V(J) is nonempty; i.e., f has a root. Conversely, let J be a proper ideal of k[x]. Since k[x] is a PID, we know J = (f) for some polynomial f. If f = 0, then V(J) is nonempty. If f is a nonzero constant, then J = k[x] is not proper, contrary to hypothesis. Thus, if f is not 0, then f is nonconstant, hence f has a root a, and since J = (f), every element of J vanishes at a. I.e., a is in V(J) so V(J) is nonempty.

[4] Let R be a commutative ring and let I and J be ideals in the polynomial ring R[x] such that I contains J. For each element f of I, assume that there is an element g of J such that f and g have the same leading coefficient (i.e., lc(f) = lc(g)), and deg(f) >= deg(g). Show that then J must equal I.

Solution: By hypothesis, I contains J, so we must show that J contains I. Assume not. Then there is an f in I that's not in J. Pick such a nonzero f of least possible degree. By hypothesis, there is a g in J with lc(f) = lc(g) and deg(f) >= deg(g). Let d = deg(f) - deg(g). Let G = xdg; thus lc(f) = lc(G) and deg(f) = deg(G), so f - G is in I but not J, but either f - G = 0 or deg(f - G) < deg(f). Since deg(f - G) < deg(f) is impossible (we already chose f to have least possible degree), we must have f - G = 0. But then f = G is in J, another contradiction. The only way out is if J contains I.

[5] (This one is for 901-902 people, but everyone of course can try it.) Hilbert was interested in finite generation for rings. The case he was interested in involved a subring R = k[g1, g2, ...] of S = k[x1, ... ,xn] generated by forms g1, g2, ..., etc., in S of positive degree. This problem shows why this is related to R being Noetherian (i.e., its ideals are finitely generated). So let I = (g1, g2, ...) be the ideal in R generated by the forms. Show that the following are equivalent:
• (a) R is finitely generated.
• (b) R is a Noetherian ring.
• (c) I is finitely generated.
Solution:
• (a) implies (b): If R is finitely generated then R = k[g1, ..., gn] for some n, so there is a surjective homomorphism from S = k[x1, ... ,xn] to R defined by sending xi to gi for each i. Thus R is a quotient of S, and we know that a quotient of a Noetherian ring is Noetherian. Hence R is Noetherian.
• (b) implies (c): All ideals in a Noetherian ring are finitely generated.
• (c) implies (a): Since I is finitely generated, there must be a finite subset of the gi which generate I. After renumbering, we may assume that I = (g1, ..., gr) for some r. I claim that k[g1, ..., gr] = R. Certainly R contains k[g1, ..., gr], so we need to verify the reverse inclusion. Since the forms gi all have positive degree, we know that every element of R of degree 0 is in k[g1, ..., gr]. Since R is generated by forms it's enough to show that every element of R of each degree is in k[g1, ..., gr]. Suppose all forms in R of degree t or less are in k[g1, ..., gr]; let h be a form in R of degree t+1. Write h = f1g1 + ... + frgr for some elements fj of R. Since h and all gj are homogeneous, we may assume that fj = 0 whenever deg(h) < deg(gi), and that otherwise deg(fj) = deg(h) - deg(gj). (If not, write each fj as fj = Fj + aj, where Fj = 0 and aj = fj whenever deg(h) < deg(gi), and otherwise Fj is the sum of all terms of fj of degree deg(h) - deg(gj) and aj is what's left over. Then h = F1g1 + ... + Frgr + a1g1 + ... + argr, but no terms in any of the summands of a1g1 + ... + argr has degree deg(h), so (since h is homogeneous) a1g1 + ... + argr = 0. Thus h = F1g1 + ... + Frgr and the F satisfy the conditions we wanted in the first place.)

Now note that deg(fj) < deg(h) since deg(gj) > 0. Thus by the induction hypothesis, fj is in k[g1, ..., gr] for each j. But this means that h is in k[g1, ..., gr], as we wanted to show. Thus by induction R = k[g1, ..., gr].