[1] Let S be the ring

- (a)
- (i) Determine V(J) in
**A**(**C**^{2}). - (ii) Conclude that Rad(J) = (x,y), where (x,y) means the ideal in S generated by x and y.

- (i) Determine V(J) in
- (b)
- (i) Explicitly find an n and m such that
x
^{m}and y^{n}are in I. - (ii) Conclude that Rad(I) = (x,y), where now (x,y) means the ideal in R generated by x and y.

- (i) Explicitly find an n and m such that
x

- (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)x^{2}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 - x^{2}= x + xx - x^{2}= 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).

- (i) Claim: V(J) = {(0,0)}.
- (b)
- (i) By playing around here I found:
-(g - yf) - x(g -(y + x)f) = x
^{3}and y(g - (y + x)f) + g - (g - yf) - x(g -(y + x)f) = y^{2}. Thus x^{3}and y^{2}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).

- (i) By playing around here I found:
-(g - yf) - x(g -(y + x)f) = x

Solution: Let J be a maximal ideal. Then the Weak Nullstellensatz says that V(J) is not empty, so let a = (a

[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[x

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 = x

[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[g

- (a) R is finitely generated.
- (b) R is a Noetherian ring.
- (c) I is finitely generated.

- (a) implies (b):
If R is finitely generated then R = k[g
_{1}, ..., g_{n}] for some n, so there is a surjective homomorphism from S = k[x_{1}, ... ,x_{n}] to R defined by sending x_{i}to g_{i}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 g
_{i 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]. }