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].