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.
[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.

[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.]

[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, and deg(f) >= deg(g). Show that then J must equal 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.