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. Solution: [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: Solution: