Math 902
Solutions to Homework # 5
1. Let R ⊆ S ⊆ T be commutative rings. Prove that T is integral over R if and only if T
is integral over S and S is integral over R.
Solution: The “only if” direction is clear. Suppose T is integral over S and S is
integral over R. Let u ∈ T . Then u is a root of a polynomial of the form xn +
c1 xn−1 + · · · + c1 x + c0 ∈ S[x]. As S is integral over R, each ci is integral over R. Let
L = R[c1 , · · · , cn ]. Then L is a subring of S which is a finitely generated R-module.
As f (x) ∈ L[x] and f (u) = 0, u is integral over L. Consequently, L[u] is a finitely
generated L-module. This means that L[u] is a finitely generated R-module. Hence, u
is integral over R.
2. Let S be a finitely generated (commutative) R-algebra which is integral over R. Prove
that for each prime p of R there are only finitely many primes of S lying over p.
Solution: Let W = R \ p. Note that any prime Q of S such that Q ∩ R = p satisfies
Q ∩ W = ∅. Hence, QW ∈ Spec SW and QW ∩ RW = (Q ∩ R)W = pW . Since SW is
integral over RW , we may replace R, S, and p with RW , SW and pW . (Note that SW
is a finitely generated RW -algebra.) Thus, we may assume (R, m) is local and p = m.
As mS ∩ R = m (by lying over), we have S/mS is integral over R/m. As the primes
of S contracting to m correspond bijectively with primes of S/mS contracting to (0)
in R/m, we may replace R with R/m and assume R is a field. In this case, S is a
finitely generated algebra over a field, so S is Noetherian. As S is integral over R,
dim S = dim R = 0. Hence, S is Artinian. As Artinian rings have only finitely many
prime ideals, the result follows.
3. Let R be a domain and K its field of fractions. For each maximal ideal m of R, we
can consider Rm as a subring of K containing R. Prove that
\
R=
Rm .
m∈maxSpec R
Solution: One containment is clear. Let α ∈ K such that α ∈ ∩m Rm . Write α = rs .
Let m be a maximal ideal of R. As rs ∈ Rm , rs = bt where b, t ∈ R, t 6∈ m. This implies
(as R is a domain) that tr = sb. Thus, t ∈ ((s) :R r). Hence, ((s) :R r) 6⊂ m. Since m
is arbitrary and ((s) :R r) is an ideal, we have that ((s) :R r) = R, i.e., r ∈ (s). Hence,
α = rs ∈ R.
4. Let R be a domain. Prove that the following are equivalent:
(a) R is integrally closed.
(b) Rp is integrally closed for every p ∈ Spec R.
(c) Rm is integrally closed for every m ∈ maxSpec R.
Solution: We proved (a) implies (b) in class and (b) implies (c) is trivial. We prove
(c) implies (a). Let K be the field of fractions of R and suppose α ∈ K is integral
over R. Let m be a maximal ideal of R. Since R ⊆ Rm ⊆ K, we have that α is
integral over Rm . As Rm is integrally closed, α ∈ Rm . Since m is arbitrary, we see
that α ∈ ∩m Rm = R by Problem #3. Thus, R is integrally closed.
5. Let k be a field. A nonempty variety V ⊆ Ank is irreducible if V is not the union of two
proper subvarieties. Prove that V is irreducible if and only if I(V ) is prime.
Solution: Suppose V is irreducible and f g ∈ I(V ) for f, g ∈ k[x1 , . . . , xn ] with f, g 6∈
I(V ). Let V = V (J) for some ideal J. Let L1 = (J, f ), L2 = (J, g), and W1 = V (L1 ),
W2 = V (L2 ). As L1 ⊃ J and L2 ⊃ J, W1 ⊆ V and W2 ⊆ V . Since f 6∈ I(V ),
there exists a point P ∈ V such that f (P ) 6= 0, which implies P 6∈ W1 . Hence,
W1 6= V . Similarly, W2 6= V . However, W1 ∪ W1 = V (L1 ) ∪ V (L2 ) = V (L1 ∩ L2 ) =
V (L1 L2 ) ⊇ V (J) = V as L1 L2 ⊆ J. Thus, V is the union of two proper subvarieties,
a contradiction. Consequently, I(V ) is prime.
Conversely, suppose I(V ) is a prime ideal. Suppose V = W1 ∪ W2 for two subvarieties
W1 , W2 of V . Then I(V ) = I(W1 ∪ W2 ) = I(W1 ) ∩ I(W2 ). As I(V ) is prime, we conclude
that either I(W1 ) = I(V ) or I(W2 ) = I(V ). Since V (I(W )) = W for every variety W ,
we have that V = W1 or V = W2 . Hence V is irreducible.
6. Let k be a field and V a nonempty affine k-variety. Prove that there exists unique
irreducible subvarieties W1 , . . . , Wr such that Wi 6⊂ Wj for all i 6= j and V = W1 ∪ · · · ∪
Wr . The Wi are called the irreducible components of V . (Hint: For existence, note
that Ank satisfies DCC on varieties. [You should say why.] By choosing a nonempty
variety minimal with respect to not being the finite union of irreducible subvarieties,
one gets a contradiction. Now throw out superfluous subvarieties in this union. For
uniqueness, use Problem # 6.)
Solution: We first show that Ank satisfies DCC on subvarieties: Suppose we have
a descending chain W1 ⊇ W2 ⊇ W3 ⊇ · · · of subvarieties. Then we have I(W1 ) ⊆
I(W2 ) ⊆ I(W3 ) ⊆ · · · is an ascending chain of ideals in k[x1 , · · · , xn ]. Hence, this
chain of ideals stabilizes; i.e., there exists a t such that I(Wt ) = I(Wt+i ) for all i ≥ 0.
Applying V (−), we obtain that Wt = Wt+i for all i ≥ 0.
Now, suppose not every subvariety of Ank is a finite union of irreducible subvarieties.
Choose W minimal with respect to not being such a union. (Here we are using DCC.)
In particular, W is reducible, so W = W1 ∪ W2 where W1 , W2 are proper subvarieties of
W . But then both W1 and W2 are the union of finitely many irreducible subvarieties,
hence so is W , a contradiction.
We’ve now proved that every subvariety V of Ank can be expressed as
V = W1 ∪ · · · ∪ Wt
where Wi are irreducible subvarieties. If Wi ⊆ Wj for some i 6= j, we can remove Wi
from this union. Hence, we may assume Wi 6⊂ Wj for all i 6= j.
We now wish to show that such a decomposition is unique. Suppose now that
V = W10 ∪ · · · ∪ Ws0
where Wi0 are irreducible varieties and Wi0 6⊂ Wj0 for i 6= j. Then
I(V ) = I(W1 ) ∩ · · · ∩ I(Wt )
= I(W10 ) ∩ · · · ∩ I(Ws0 ).
Since I(W1 ) ⊇ I(W10 ) ∩ · · · ∩ I(Ws0 ) and I(W1 ) is prime (by Problem #5), we have
I(W1 ) ⊇ I(Wi0 ) for some i. For the same reason, I(Wi0 ) ⊇ I(Wj ) for some j. Thus,
I(W1 ) ⊇ I(Wj ). Applying V (−) to both sides, we get W1 ⊆ Wj , which implies j = 1.
Hence, I(W1 ) = I(Wi0 ), which gives that W1 = Wi0 . Reordering, we can assume that
W1 = W10 . Repeating with W2 , . . . , Wt , we get obtain that s = t and (after reordering),
Wi = Wi0 for all i.
7. Let k be a field and V ⊆ Ank a nonempty affine variety. Define the dimension of V
to be the supremum of integers ` ≥ 0 such that there exists a chain of irreducible
subvarieties of V :
W0 ( W1 ( · · · ( W` .
Prove that if k is algebraically closed, then dim V = dim k[x1 , . . . , xn ]/ I(V ).
Solution: Let d be the dimension of V and ` = dim k[x1 , . . . , xn ]/I(V ). Then there
exists a chain of irreducible subvarieties of V :
W0 ( W1 ( · · · ( Wd .
Then we have a chain of prime ideals of k[x1 , . . . , xn ] containing I(V ):
I(W0 ) ⊃ I(W1 ) ⊃ · · · ⊃ I(Wd ).
Each of these containments is strict, since if I(Wi ) = I(Wi+1 ) then Wi = Wi+1 . Hence,
d ≤ `.
Then there exist a chain of prime ideals containing I(V ):
P` ( P`−1 ( · · · ( P0 .
Hence, we get a chain of subvarieties of V = V (I(V )):
V (P` ) ⊃ V (P`−1 ⊃ · · · ⊃ V (P0 ).
√
As k is algebraically closed, we have by the Nullstellensatz that I(V (Pi )) = Pi = Pi
for each i. Hence, by Problem # 5, V (Pi ) is irreducible and also V (Pi ) ) V (Pi−1 ) for
all i. Thus, ` ≤ d and hence ` = d.