2
$\begingroup$

I'm currently reading Rotman's An Introduction to Homological Algebra (2nd edition), and on page 188 in the proof of Theorem 4.66 (Every left Artinian ring is semiperfect), I ran across the claim:

If $R$ is left Artinian, and $J(R)$ is the Jacobson radical of $R$, then $R/J(R)$ is left semisimple.

Now, it's quite possible that I've run across this claim earlier in the text accompanied by a citation likely of either Lam's A First Course in Noncommutative Rings or Rotman's own Advanced Modern Algebra (the books Rotman usually cites for algebra facts), but there is no citation for the fact at this point in the text.

However, I don't have a copy of either of these books, so I decided to try to reconstruct a proof.

It suffices to prove that if $R$ is left Artinian and semiprimitive ($J(R)=0$), then it is left semisimple.

Attempted proof:

Consider the set of finite intersections of maximal left ideals. Since $R$ is left Artinian, this collection has a minimal element, $I=M_1\cap \cdots \cap M_n$. Then $I=J(R)=0$, since if $m\in I$ with $m\ne 0$, then since the intersection of all maximal ideals is $J(R)=0$, there is a maximal ideal $M_0$ with $m\not\in M_0$, so $I\cap M_0 \subsetneq I$, contradicting the minimality of $I$.

Thus let $M_1,\ldots,M_n$ be maximal left ideals such that $M_1\cap M_2\cap\cdots \cap M_n=0$.

Now consider the map $$\phi : R\to \bigoplus_{i=1}^n R/M_i$$ defined by $r\mapsto (r+M_i)$.

The kernel of this map is $\bigcap_{i=1}^n M_i=0$, so this map is injective. Hence, we have realized $R$ as a submodule of the semisimple module $\bigoplus_{i=1}^n R/M_i$.

Thus we have realized $R$ as a submodule of a semisimple (left) $R$-module, so $R$ is left semisimple itself (Corollary 4.2 in Rotman Hom. Alg.). $\blacksquare$

Questions

  1. Is this proof correct or am I missing something?
  2. My first thought was to take $R\to \prod_M R/M$ with $M$ running over all maximal left ideals, and I hoped that the image of $R$ would end up in the direct sum, but I realized that I couldn't guarantee that this was going to be the case unless there were finitely many maximal ideals. In the commutative case it's not hard to see immediately that an Artinian ring has finitely many maximal ideals, since we have the lemma

    If $Q \supseteq \bigcap_{i=1}^n P_i$, then $Q\supseteq P_i$ for some $i$. (Proof: Otherwise take $p_i\in P_i\setminus Q$ and consider $\prod_i p_i$.)

    The contrapositive of this says that if there were infinitely many maximal ideals, their successive finite intersections wouldn't stabilize. Is there a similar argument to prove that there are only finitely many maximal ideals in the noncommutative case? Naturally, this follows from semisimplicity, but I was wondering if there's a way to see this a priori.

Note: I wrote the bulk of this yesterday, before finding a copy of Advanced Modern Algebra online. It does in fact contain a (different) proof of this proposition. (Theorem 8.45 in the second edition) Since the proof is different, question (1) remains valid, and I'm still curious about question (2).

Edit note: I am editing the writing at several points to make my thoughts, intentions and argument more clear, following the advice of rschweib's answer. This note is for future readers who may therefore be puzzled by the parts of rschweib's answer commenting on certain things in the question that were poorly expressed.

$\endgroup$
1
$\begingroup$

The proof is correct, although it would be preferable to explain why the zero ideal is a finite intersection of maximal right ideals a little more carefully. It is true that in an artinian poset of one sided ideals, a collection intersecting to zero must have a finite subset intersecting to zero, but it is not really done justice by saying “the set had a minimal element which is obviously (read “necessarily”) zero.”

My first thought was to take $R→⨁_M R/M$ [...] with $M$ running over all maximal left ideals, but I realized that I couldn't guarantee that this was a direct sum a priori.

Well, you can make that sum and it will be direct, nothing prevents you from doing so. But yes, the map you want to do (sending $r$ to its image in the quotient) doesn’t work. You would necessarily be working in $\prod_M R/M$ instead. Reducing it to a finite product makes it equivalent to a direct sum, and that was a very important reduction.

Is there a similar argument to prove that there are only finitely many maximal ideals in the noncommutative case?

Yes, the same argument (with noncommutative prime ideals) works. If a prime ideal contains a finite intersection of prime ideals, it necessarily contains their product, whence it contains at least one of the terms the product.

$\endgroup$
  • $\begingroup$ Thanks! Yes, both criticisms of my writing are on point. I should've been more explicit about why the minimal element had to be zero. Also, yes I did mean that I wanted to take the map $R\to \prod_M R/M$ and have it end up in the direct sum, but I realized that that wasn't going to be true unless I could get it down to finitely many modules. I'm particularly grateful for the suggestion to consider noncommutative prime ideals. I'm generally quite unfamiliar with noncommutative rings, so I wasn't even aware of the generalization. That looks very interesting :) Thanks! $\endgroup$ – jgon Jan 2 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.