M953 Homework 2 (Click
Due Friday, February 8, 2002
 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.
 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.
- (i) Determine V(J) in
- (ii) Conclude that Rad(J) = (x,y), where (x,y) means the ideal
in S generated by x and y.
- (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.
 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.]
 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.
 (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.