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Math 6421
Fall 2009
HW assignment #1 solutions
1. Let k be an algebraically closed field, and let Z ⊆ A3k be the set Z =
{(t, t2 , t3 ) | t ∈ k}. Find generators for the ideal I(Z) in k[x, y, z], and
show that the ring A(Z) = k[x, y, z]/I(Z) is isomorphic to a polynomial
ring in one variable over k. (Z is called the twisted cubic curve in A3 .)
Solution:
Let J = (y − x2 , z − x3 ). It is clear that J ⊆ I(Z), since y − x2
and z − x3 both vanish identically on Z. For the other inclusion, take
f = f (x, y, z) ∈ I(Z), and let g = g(x, y, z) = f (x, x2 , x3 ) ∈ k[x] ⊂
k[x, y, z]. Then f − g ∈ J, since y ≡ x2 and z ≡ x3 (mod J). Since
f ∈ I(Z), we have f (t, t2 , t3 ) = 0 for all t ∈ k. Thus f (x, x2 , x3 ) is the
zero polynomial in k[x], since it vanishes at every t ∈ k. (Note that k
is infinite.) In other words, g = 0, and thus f ∈ J as desired.
Now define φ : k[x, y, z] → k[t] by sending x to t, y to t2 , and z
to t3 . Clearly φ is surjective, and by definition ker(φ) = I(Z), since
f ∈ ker(φ) iff φ(f ) = f (t, t2 , t3 ) = 0 in k[t] iff f vanishes identically on
Z. Thus A(Z) = k[x, y, z]/I(Z) ∼
= k[t].
Remark: It is not obvious that J is the kernel of φ, so if your proof of
J = I(Z) was based on this observation, make sure that you justify it.
2. Let k be an algebraically closed field, let A be a√finitely generated kalgebra, and let I be an ideal in A. Prove that I is the intersection
of all maximal ideals of A containing I. (Compare with Exercise 2(b)
below.)
Solution:
Every finitely generated k-algebra A has the form k[x1 , . . . , xn ]/J for
some ideal J. Also, ideals in k[x1 , . . . , xn ]/J correspond to ideals in
k[x1 , . . . , xn ] containing J, and this correspondence preserves maximal
ideals and radicals. So we may assume without loss of generality that
A = k[x1 , . . . , xn ].
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Now, it is geometrically obvious that the algebraic set V(I) ⊂ Ank is
the union of all points contained in it:
[
V(I) =
P .
P ∈V(I)
Taking I√of both sides, we obtain (using Hilbert’s Nullstellensatz to
see that I = I(V(I)) and that points in Ank correspond to maximal
ideals in k[x1 , . . . , xn ]):
\
√
I = I(V(I)) =
m.
m⊇I
3. If Y ⊆ An is any subset, prove that V (I(Y )) = Y (the closure of Y in
the Zariski topology).
Solution:
From the definitions of V and I, it is rather clear that we always have
Y ⊆ V(I(Y )).
Now assume V is an arbitrary algebraic set containing Y : V ⊇ Y .
Then I(V ) ⊆ I(Y ) and V(I(V )) ⊇ V(I(Y )). Since V is an algebraic
set, V = V(I(V )). So V ⊇ V(I(Y )). Therefore V(I(Y )) is the smallest
algebraic set containing Y .
4. Let X be a topological space. Prove that the following are equivalent:
(a) X is irreducible (i.e., X cannot be written as X = X1 ∪ X2 with
X1 , X2 proper closed subsets of X).
(b) Any two non-empty open subsets of X intersect.
(c) Every non-empty open subset of X is dense in X.
Solution:
(a) implies (b):
open subsets of X.
T Let U1 and U2 be two
S non-empty
c
c
c
Suppose U2 U2 = ∅. Then X = U1 U2 , and U1 and U2c are proper
closed subsets of X, a contradiction.
(b) implies (c): Recall that a subset of X is dense in X if and only if it
meets all nonempty open subsets of X. Thus it follows from (b) that
all non-empty open subsets of X is dense in X.
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(c) implies (a): Suppose for contradiction
that X is not irreducible.
S
Then we can write X as X = X1 X2 with X1 and X2 proper closed
subsets of X. Then X1c ⊆ X2 , which means that X1c is a nonempty
open subset of X whose closure is not X and thus X1c is not dense in
X.
5. Let X be a Noetherian topological space.
(a) Prove that X is quasicompact, i.e., every open cover has a finite
subcover.
(b) If X is Hausdorff, prove that X is a finite set with the discrete
topology.
(c) Prove or find a counterexample: X has finite Krull dimension.
Solution:
(a) Suppose for the sake of contradiction that there exists an open cover
{Uα }α∈A of X with no finite subcovers. Then we can find a sequence
S
of non-empty open T
sets {Uα1 , · · · , Uαn , · · · } such that Uαi+1 * j6i Uαj
for all i. Let Yi = j6i Uαc j . Then Y1 ⊃ Y2 ⊃ · · · does not satisfy the
descending chain condition for closed subsets.
(b) Recall that we say X has the discrete topology if every subset of X
is open (closed). Now let X be a Noetherian Hausdorff space. Suppose
X has infinitely many points. Then pick two distinct points x1 and y1 ,
and we can find disjoint neighborhoods U1 3 x1 and V1 3 y1 . Either
X \ U1 or X \ V1 is infinite. WLOG, we assume X \ U1 is infinite. Let
U10 = U1 . Now X \ U1 is also Hausdorff and we can find U2 ⊆ XS\ U1
which is non-empty and open relative to X \ U1 such that X \ (U1T U2 )
is infinite. Then there exists an open subset U20 of X such that U20 (X \
U1 ) = U2 . In this way, we can continue
S
S to find Ui such that
S UiSis nonempty, open relative to X \ (U1 · · · Ui−1 ) and X \ (U1 · · · Ui ) is
infinite.
Correspondingly,
there exist open subsets
Ui0 S
of X such that
T
S
S
S
0
0
Ui (X \ (U1 · · · Ui−1 )) = Ui . Let Zi = U1 · · · Ui0 . Then by
construction, Ui0 * Zi−1 , which means the ascending chain of open sets
Z1 ⊂ Z2 ⊂ · · · never stabilizes, contradicting the assumption that X
is Noetherian. Thus X is finite. And since every point of a Hausdorff
space is closed, X has the discrete topology.
(c) A Noetherian space can have infinite Krull dimension. Let us build
an infinite dimensional Noetherian space. Let X be the unit interval
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[0, 1]. Let X and ∅ be closed. Let the subintervals [1/n, 1] be closed
for n ∈ Z+ . Then X is a Noetherian topological space and all closed
subsets are irreducible. But now we get an ascending chain of distinct
irreducible closed subsets of X as [1, 1] ⊂ [1/2, 1] ⊂ · · · ⊂ [1/n, 1] ⊂ · · · .
It follows that X has infinite Krull dimension.
6. Let A be an integral domain with field of fractions K. Prove that the
integral closure of A in K is the intersection of all valuation rings of K
containing A. [Hint: Use the Extension Theorem.]
Solution:
It suffices to show that x ∈ K is integral over A iff x belongs to every
valuation ring of K containing A.
Suppose first that x is integral over A, and let B be a valuation ring
of K containing A. Then either x ∈ B or x−1 ∈ B. If x ∈ B, we are
done. If x−1 ∈ B, then by the integrality of x we have
xn + a1 xn−1 + · · · + an−1 x + an = 0
with ai ∈ A, and thus
x = −(a1 + a2 x−1 + · · · + an (x−1 )n−1 ) ∈ B
as desired.
Conversely, suppose x ∈ K is not integral over A, and let A0 = A[x−1 ].
Then x 6∈ A0 by an argument similar to the one above. In particular,
x−1 is a non-unit in A0 , so is contained in some maximal ideal m0 of A0 .
Let k be an algebraic closure of A0 /m0 . The natural map φ from A to
A0 to A0 /m to k sends x−1 to 0, and by the Extension Theorem, φ can
be extended to a homomorphism Φ : B → k for some valuation ring B
of K containing A. As Φ(x−1 ) = 0, we must have x 6∈ B.
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Supplementary commutative algebra exercises
1. (?) (Localization) Let A be a commutative ring with identity, and S a
multiplicative subset of A.
(a) If I is an ideal of A, prove that S −1 I (the image of I under the
natural homomorphism from A to S −1 A) is a proper ideal iff I ∩
S = ∅.
(b) Prove that every ideal of S −1 A has the form S −1 I for some ideal
I of A.
(c) Prove that there is a bijection between prime ideals of S −1 A and
prime ideals of A disjoint from S.
(d) If S = A\p with p a prime ideal of A, prove that Ap := S −1 A is a
local ring, i.e., that it has a unique maximal ideal.
Solution: See, e.g., pages 708,709, and 718 of [DF04] or pages 41,42
of [AM69].
2. (?) (Radical of an ideal) Let A be a commutative ring with identity.
(a) If f ∈ A is not nilpotent, prove that there is a prime ideal of A
which does not contain f . [Hint: Since S = {1, f, f 2 , . . .} does
not contain 0, the localization S −1 A = Af is not the zero ring, so
it has a maximal (and hence prime) ideal.]
√
(b) If I is an ideal in A, show that I is the intersection of all prime
ideals of A containing I. [Hint: The image of f ∈ A in A/I is
nilpotent iff f m ∈ I for some m ≥ 1.]
Solution: See, e.g., page 71 of [Eis95]. For a proof of (b) without
localization, see e.g. page 9 of [AM69].
3. (?) (Integral closure)
(a) Let A ⊆ B be rings, let C be the integral closure of A in B, and let
S be a multiplicative subset of A. Prove that S −1 C is the integral
closure of S −1 A in S −1 B. In particular, every localization of an
integrally closed domain is again integrally closed.
(b) Prove that a UFD is integrally closed.
Solution: See page 61 of [AM69] and page 693 of [DF04].
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References
[AM69] M. F. Atiyah and I. G. Macdonald. Introduction to commutative
algebra. Addison-Wesley Publishing Co., Reading, Mass.-LondonDon Mills, Ont., 1969.
[DF04] D. S. Dummit and R. M. Foote. Abstract algebra. John Wiley &
Sons Inc., Hoboken, NJ, third edition, 2004.
[Eis95] D Eisenbud. Commutative algebra, volume 150 of Graduate Texts in
Mathematics. Springer-Verlag, New York, 1995. With a view toward
algebraic geometry.
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