MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #6

JOHN VOIGHT

Problem 1. Let R be a ring. Let N = rad R = ⋂_{𝔪} 𝔪 be the intersection of all maximal left ideals of R.

(a) Show that NE = 0 for every simple R-module E. Show that N is a twosided ideal.

(b) Show that rad(R/N) = 0.

Solution. By the beginning of (XVII §6), there is a bijection between maximal left ideals and simple R-modules (up to isomorphism) and E ≅ R/M for a maximal left ideal M. Already, then, N ⊂ M = Ann E, so NE = 0. Moreover, since N(R/M) = 0 for all maximal left ideals M, NR ⊂ M for all M and therefore NR ⊂ ⋂_{M} M = N, so N is also a right ideal.

For (b), note that every maximal ideal of R/N is the image of a maximal ideal M of R containing N (under the natural surjection R → R/N). But every maximal ideal contains N, so rad(R/N) is the image of rad(R) = N in R/N, i.e. rad(R/N) = 0.

Problem 2. A ring is (left) artinian if every descending sequence of left ideals stabilizes.

(a) Show that a finite-dimensional algebra A over a field k is artinian.

(b) If R is artinian, show that every nonzero left ideal contains a simple left ideal.

(c) If R is artinian, show that every nonempty set of left ideals S contains a minimal left ideal.

Solution. A left ideal of A is in particular a finite-dimensional k-vector space, so for any sequence J_{1} ⊃ J_{2} ⊃ … we have dim_{k} A ≥ dim_{k} J_{1} ≥ … and this can only have finitely many strict inequalities; therefore the original sequence of ideals stabilizes.

For (b), let J = J_{1} be any nonzero left ideal of R. If J itself is simple, we are done; otherwise it properly contains a nonzero left ideal J_{2}. Continuing in this fashion, we obtain J = J_{1} ⊋ J_{2} ⊋ … which stabilizes as R is artinian. The final stable factor J_{n} ≠ 0 is a simple left ideal of J.

Finally, for (c) let J_{1} ∈ S; if J_{1} is not minimal, there exists J_{2} ∈ S such that J_{1} ⊋ J_{2}. Continuing in this way we obtain a descending chain of left ideals and since R is artinian, this procedure must eventually terminate and J_{n} is a minimal ideal in S.

Date: March 18, 2003.

XVII: 1–7, 9, 10, 13.

Problem 3. Let R be artinian. Show that rad R = 0 if and only if R is semisimple.

Solution. We must assume that R is not the zero ring since by definition the zero ring is not semisimple (§4).

Since R is artinian, we claim that we can write rad R = ⋂_{i=1}^{t} M_{i} for a finite set of maximal (left) ideals M_{1}, …, M_{t}. Otherwise, there would exist a descending chain M_{1} ⊋ M_{1} ∩ M_{2} ⊋ … of left ideals of R.

Suppose that rad R = 0. Then the map R → ⨁_{i} R/M_{i} is injective, since its kernel is ⋂_{i} M_{i} = rad R = 0. But each R/M_{i} is simple, since each M_{i} is maximal; therefore R is (isomorphic to) a submodule of a semisimple module, so R is semisimple.

If R is semisimple, write R ≅ ⨁_{i} E_{i} an isomorphism of R-modules with E_{i} simple. Then M_{i} = Ann E_{i} is a maximal left ideal of R and ⋂_{i} Ann E_{i} = ⋂_{i} M_{i} = 0. But rad R ⊂ ⋂_{i} M_{i} = 0, so rad R = 0. Note this does not require that R be artinian!

Problem 4 (Nakayama’s Lemma). Let R be a ring and M a finitely generated module. Let N = rad R. If NM = M, show that M = 0.

Solution. Let m_{1}, …, m_{r} be a minimal generating set for M (i.e. no smaller subset generates M). Then there exist n_{1}, …, n_{r} ∈ N such that n_{1}m_{1} + ⋯ + n_{r}m_{r} = m_{r}, so (1 − n_{r})m_{r} = n_{1}m_{1} + ⋯ + n_{r−1}m_{r−1}. We know 1 − n_{1} must be left invertible (otherwise 1−n_{1} is contained in some maximal left ideal and so too would n_{1} + (1 − n_{1}) = 1)—but then m_{1}, …, m_{r−1} generate M, contradicting minimality. Therefore M had no such minimal generating set, i.e. M = 0.

Problem 5.

(a) Let J be a two-sided nilpotent ideal of R. Show that J is contained in the (Jacobson) radical.

(b) Conversely, assume that R is Artinian. Show that its Jacobson radical is nilpotent, i.e., that there exists an integer r ≥ 1 such that N^{r} = 0. [Hint: Consider the descending sequence of powers N^{r}, and apply Nakayama to a minimal finitely generated left ideal L ⊂ N^{∞} such that N^{∞}L ≠ 0.]

Solution. For (a), suppose that x ∈ J has x ∉ M for some maximal ideal M. Then Rx + M = R, so there exists an a ∈ R and m ∈ M such that ax + m = 1. But ax ∈ J is nilpotent, so say (ax)^{n} = 0. Then

(1 + ⋯ + (ax)^{n−1})(1 − ax) = 1 − (ax)^{n} = 1

so m = 1 − ax is left invertible, a contradiction since m ∈ M. Hence no such x exists, and J ⊂ rad R.

For (b), let N = rad R; then N ⊃ N^{2} ⊃ … is a descending chain of left ideals, so it stabilizes, say the limit is N^{r} = N^{r+1} = . . .. Suppose that N^{r} ≠ 0: then in particular, N^{r}N = N^{r+1} = N^{r} ≠ 0. Let S be the set of left ideals L such that N^{r}L ≠ 0. We have shown that N ∈ S, so S ≠ ∅, so by Exercise 2(c), there is a minimal element L ∈ S. By minimality, we know that L is finitely generated (if N^{r}L ≠ 0 there exists x ∈ L such that N^{r}x ≠ 0, so Rx ∈ S and hence L = Rx). By Nakayama’s lemma, since N(N^{r}L) = N^{r}L, we must have N^{r}L = 0, a contradiction. Therefore N^{r} = 0.

Problem 6. Let R be a semisimple commutative ring. Show that R is a direct product of fields.

Solution. By Theorem 4.3, every semisimple ring is the (finite) product of simple rings. A commutative simple ring k must be a field: the zero ring is not simple, and if a ∈ k is nonzero, then (a) is a (two-sided) ideal hence by Theorem 5.2, (a) = k, so a ∈ k^{∗} and k is a field.

Problem 7. Let R be a finite-dimensional commutative algebra over a field k. If R has no nilpotent element ≠ 0, show that R is semisimple.

Solution. R is Artinian by Exercise 2(a), and rad R is nilpotent by Exercise 5(b); by hypothesis, then, rad R = 0, so by Exercise 3, R is semisimple.

Problem 10. Let E be a finite-dimensional vector space over a field k. Let A ∈ End_{k}(E). Show that the k-algebra R generated by A is semisimple if and only if its minimal polynomial has no factors of multiplicity > 1 over k.

Solution. Let m(t) = ∏_{i=1}^{r} p_{i}(t)^{e_{i}} be the minimal polynomial of A written as the product of irreducibles. Then by the Chinese remainder theorem,

R = k[t]/(m(t)) ≅ ∏_{i=1}^{r} k[t]/(p_{i}(t)^{e_{i}}).

The k-algebra R is commutative, so by Exercise 6, R is semisimple (if and) only if R is a direct product of fields. But R is a product of fields if and only if e_{i} = 1 for each i, for otherwise one factor has a nilpotent element, which is the result.