Math 902
Solutions to Homework # 2
Throughout R denotes a commutative ring with identity.
1. Prove that R has nontrivial idempotents if and only if R ∼
= S × T , where S and T are
(nonzero) rings with identity.
Solution: Suppose R has a nontrivial idempotent e. Then 1 − e is also a nontrivial
idempotent. It is easy to see that the ideal Re is itself a nonzero ring with multiplicative
identity e. Similarly for R(1 − e). Define a map f : R → Re × R(1 − e) by f (r) =
(re, r(1 − e)). It is easily seen that f is a ring homomorphism. If f (r) = (0, 0) then
re = r(1 − e) = 0, which gives r = 0. Hence, f is injective. Given (re, s(1 − e)) ∈
Re × R(1 − e), let u = re + s(1 − e). Then f (u) = (re, s(1 − e)). Thus, f is surjective.
Conversely, if R ∼
= S × T , the elements of R corresponding to (1, 0) and (0, 1) under
the isomorphism are nontrivial idempotents.
2. Find Spec Z/(100)Z and Spec R[x]/(x4 − 1).
Solution: Recall that the prime ideals of a quotient ring R/I are of the form P/I
where P is a prime ideal of R containing I. The prime ideals of Z containing the ideal
(100)Z correspond to the prime divisors of 100: (2)Z and (5)Z. Thus, Spec Z/(100)Z =
{(2)Z/(100)Z, (5)Z/(100)Z}.
The prime ideals of R[x] containing (x4 − 1) correspond to the irreducible factors of
x4 − 1: (x − 1), (x + 1), and (x2 + 1). Taking the quotient of these prime ideals by
(x4 − 1) gives the set of primes of R[x]/(x4 − 1).
3. Recall that a topological space X is said to be T1 if given x, y ∈ X, x 6= y, there exists
an open set U such that x ∈ U and y 6∈ U . Prove that Spec R is T1 if and only if
dim R = 0.
Solution: Suppose Spec R is T1 , but dim R > 0. Then there exist prime ideals
p ) q. Since Spec R is T1 , there exists an open set U such that p ∈ U , q 6∈ U . Let
U = Spec R \ V(I). Then p 6⊃ I but q ⊃ I, a contradiction.
Conversely, suppose dim R = 0. Let p 6= q be prime ideals, and U = Spec R \ V(q).
Then p ∈ U and q 6∈ U , Hence, Spec R is T1 .
4. A topological space is called Noetherian if it satisfies√the ascending chain condition
(ACC) on open sets. An ideal I is called radical if I = I. Prove that Spec R is
Noetherian if and only if R satisfies ACC on radical ideals.
√
√
Solution: First observe that for ideals I and J, V(I) ⊆ V(J) if and only if I ⊇ J.
This is a consequence of Krull’s theorem which says that the radical of an ideal is the
intersection of all prime ideals containing it. Note also that Spec R is Noetherian if
and only if it satisfies DCC on√closed sets, i.e., subsets of the form V(I), where I is a
radical ideal (since V(I) = V( I)). The result is now clear from these observations.
5. Let R be a Noetherian ring and I 6= R an ideal. Prove that the set of minimal primes
over I is finite. (Hint: If not, choose I maximal with respect to the property that
I 6= R and I has infinitely primes minimal over it. Derive a contradiction.)
Solution: As in the hint, assume the result is false and let I]neqR be an ideal maximal
with respect to the property that of having infinitely many minimal primes. (This is
possible as R is Noetherian, not by Zorn’s lemma.) Then clearly I is not a prime ideal,
so there exist elements a, b ∈ R such that ab ∈ I but a 6∈ I, b 6∈ I. Let J = (I, a)
and K = (I, b). Then J and K properly contain I, but JK ⊆ I. Note that any prime
containing I must contain either J or K. In particular, any prime minimal over I is
minimal over either J or K. But J and K each have only finitely many minimal primes
(by choice of I), a contradiction.
6. Let R be a Noetherian ring. Prove that Spec R is Hausdorff if and only if dim R = 0.
Solution: Is Spec R is Hausdorff then it is T1 . Hence, dim R = 0 by Problem #3.
Suppose dim R = 0. As R is Noetherian, R has only finitely many minimal prime
ideals by Problem #5. Hence, Spec R is a finite set. We claim that Spec R has the
discreet topology. As Spec R is finite, it is enough to show that every point is closed.
Let p ∈ Spec R. Then {p} = V(p) (as there are no containment relations among the
prime ideals). Hence, Spec R is discrete.
7. Let f : R → S be a nonzero homomorphism of commutative rings with identity.
(a) For q ∈ Spec S, prove that f −1 (q) := {r ∈ R | f (r) ∈ q} is a prime ideal of R.
(b) Show that the map f ∗ : Spec S → Spec R given by f ∗ (q) = f −1 (q) is continuous.
Solution: For part (a), let q ∈ Spec S and suppose ab ∈ f −1 (q). Then f (a)f (b) =
f (ab) ∈ q. Thus, f (a) ∈ q or f (b) ∈ q and consequently a ∈ f −1 (q) or b ∈ f −1 (q).
For part (b), let V (I) be a closed set of Spec R. It sufflices to show (f ∗ )−1 (V(I)) =
V(IS). Let q ∈ (f ∗ )−1 (V(I)). Then f ∗ (q) = f −1 (q) ⊇ I. Thus, q ⊇ IS. Conversely,
suppose q ∈ V(IS). Then f ∗ (q) = f −1 (q) ⊇ f −1 (I) ⊇ I, so f ∗ (q) ∈ V(I). Thus,
q ∈ (f ∗ )−1 (V(I))