6
$\begingroup$

Just to be consistent with the terminology, let me define the words I'm using.

A module $M$ is simple if it has no proper non-zero submodules, and semi-simple if it can be written as a direct sum $M = \bigoplus_i A_i$ where the $A_i$'s are simple submodules.

A ring $R$ is semi-simple if the left $R$-module $_RR$ is semi-simple. This means that $R$ is a direct sum of "minimal" ideals. I want to prove that such rings are also Dedekind-finite, i.e. if $a,b\in R$ and $ab = 1$, then $ba = 1$. I want to do this without invoking Artin-Wedderburn. (EDIT: We have "semi-simple $\implies$ Noetherian $\implies$ Dedekind-finite", but I don't know how to prove the first implication without using composition series. It seems to me that my problem can be solved without using the fact that semi-simple rings are Noetherian.)

So far, I can show that a semi-simple ring $R$ is a direct sum of finitely many minimal ideals: if $R = \bigoplus_\lambda A_\lambda$ where each $A_\lambda$ is a minimal ideal of $R$, then write $1 = e_1 + \cdots + e_n$ where $e_i \in A_{\lambda_i}$. It follows that $r = re_1 + \cdots + re_n$ for any $r \in R$, so $R = \bigoplus_{i=1}^n A_{\lambda_i}$. Let's abbreviate the notation by setting $A_i := A_{\lambda_i}$.

Alright, now suppose $a,b\in R$ are such that $ab = 1$. We want to show that $ba = 1$. I have attempted to solve the problem in the case that $a \in A_i$ for some $i$. First, we get $ab = 1 \implies bab = b \implies (ba - 1)b = 0$. Since $b\neq 0$, we'd be done if $R$ is a domain. Even if $R$ is not a domain, I think we can show that we must have $ba - 1 = 0$, although my approach doesn't work.

Define an $R$-module homomorphism $F: R \rightarrow R$ by $F(r):= rb$. Suppose that $a\in A_i$ for some $i$. Then $F$ is non-zero on $A_i$, since $F(a) = ab = 1 \neq 0$. (Here's where we use that $a \in A_i$.) If we restrict $F$ to $A_i$, the kernel of this restriction is proper (since $F$ is non-zero) and therefore the kernel must be $\{0\}$ (because $A_i$ is minimal). Therefore the restriction of $F$ to $A_i$ is one-to-one.

Note that $F(ba - 1) = (ba - 1)b = 0$ as mentioned above. Since $F$ is one-to-one, we must have $ba - 1 = 0$, as required.

Problem with this last argument: we only know that $F$ is one-to-one on $A_i$, but we don't know that $ba - 1 \in A_i$. In fact, if $ba - 1 \in A_i$, then $1 = ba + -(ba - 1) \in A_i$ since $a \in A_i$. Thus $A_i$ is not proper, so $_RR = {_R}A_i$ is a simple $R$-module, which means that $R$ is a division ring (it's easy to prove that "$_RR$ simple $\implies$ $R$ is a division ring"). Since division rings are clearly Dedekind-finite, the problem is trivial in this case.

Anyway, so my attempt using a homomorphism $F$ doesn't really work. But it might be possible to salvage the approach by modifying it slightly. Even then, this would only give the result when $a \in A_i$ for some $i$, but hopefully this special case can be used to prove the more general claim.

Reminder that I want to prove that semi-simple rings are Dedekind-finite without using the Artin-Wedderburn theorem.

Cheers!

$\endgroup$
  • $\begingroup$ I prepended a new argument. You'll have to let me know if you're not willing to use the fact that the number of minimal ideals in the decomposition of $R$ is invariant. $\endgroup$ – rschwieb May 23 '13 at 11:07
3
$\begingroup$

Prepended: OK, here is an elementary argument based only on the simple submodules. We begin by assuming $R$ is a direct sum of $n$ minimal right ideals, and we know that $n$ is unique.

If $ab=1$, then the homomorphism $x\mapsto bx$ from $R\to R$ is injective. Thus $bR\cong R$ as $R$ modules. Find a complement $C$ to $bR$ so that $R=bR\oplus C$. If $ba\neq 1$, then $bR\neq R$ and $C\neq \{0\}$.

But look: $R\cong R\oplus C$, and since $C$ is nonzero it contributes at least one simple submodule to the sum. But that would mean that the right hand side of the equation is a sum of more than $n$ simple modules, whereas the left is the sum of $n$ simple modules. This isn't possible, so $bR$ must be $R$, showing that $ba=1$.


As soon as you have convinced yourself $R$ is a direct sum of finintely many minimal right ideals, then you immediately have that $R$ has a finite composition series, hence it is right Artinian (and right Noetherian.)

I think it is even easier to use the Artinian property than the Noetherian property.

Suppose $ab=1$. Clearly $b$ is left invertible. Consider the descending chain $bR\supseteq b^2R\supseteq\dots$. Since semisimple rings are right (and left) Artinian, this chain must stabilize, say at $b^nR=b^{n+1}R$. This implies $b^n=b^{n+1}r$ for some $r\in R$. Multiplying on the left by $a$ $n$ times, you get $1=br$, showing that $b$ is right invertible.

Thus $a=abr=r$, and so $ba=1$.


Note: There is a way to prove a semisimple ring is Dedekind finite independently of the chain conditions. Since semisimple rings are self-injective on both sides, they are Dedekind finite. Unfortunately, this route seems even more complicated than the methods above, but still, it is interesting to see that you can get Dedekind finiteness "two different ways" for semisimple rings.

$\endgroup$
  • $\begingroup$ Not willing to assume facts about composition series either, huh :P Well I'll keep thinking... Your self-handicap certainly robs us of all the straightforward, commonsense approaches... $\endgroup$ – rschwieb May 23 '13 at 1:38
  • $\begingroup$ ^ Yeah, my bad. I'm just trying to create a self-contained exposition of this material, so I don't want to rely on stuff outside of what I've done so far. But yeah -- I'd better tell you what I've done so far then, huh? I know that "semi-simple" and "semi-primitive" (= admits a faithful semi-simple module) are equivalent for Artinian rings. I also have the Jacobson Density Theorem, so that any primitive Artinian ring is of isomorphic to a ring of the form $\matbb{M}_n (\Delta)$, where $n \geq 0$ and $\Delta$ is a division ring. $\endgroup$ – Ehsaan May 23 '13 at 19:13
  • $\begingroup$ Also, it's important to note that every time I say "ideal", I mean "left ideal" -- the ring $R$ is if the left $R$-module $_RR$ is simple, etc. Anyway, thanks for your input so far! $\endgroup$ – Ehsaan May 23 '13 at 19:18
  • $\begingroup$ @Ehsaan If you are just concerned about overhead learning cost, then I would say that learning that a module has finite composition length iff it is Artinian and Noetherian is very worthwhile, and it is easy to apply in this situation to show that a semisimple ring must be (left and right) Artinian. It really seems economical to me. $\endgroup$ – rschwieb May 23 '13 at 19:30
  • $\begingroup$ @Ehsaan A small challenge for you, considering you are learning about semiprimitive and right primitive rings: give an example of a right primitive ring which isn't Dedekind fininte. $\endgroup$ – rschwieb May 23 '13 at 19:34
2
$\begingroup$

As $R$ is semi-simple, $R$ is noetherian. If $ab = 1$, the map $F \colon R \to R$, $F(r) = rb$ is onto. We will show that it is an isomorphism. Let $I_j = \ker F^j$. As $R$ is noetherian, there is an $j$ such that $I_{j+1} = I_j$. Now suppose $F(a) = 0$, then $a = F^j(a')$ for some $a' \in R$, as $R$ is onto, so $$ 0 = F(a) = F^{j+1}(a') $$ hence $a' \in I_{j+1} = I_j$, so $a = F^j(a') = 0$. Hence $F$ is one-to-one. As you wrote $F(ba - 1) = 0$ now gives $ba = 1$ and $R$ is Dedekind-finite.

$\endgroup$
  • $\begingroup$ Hmm ... I've also seen this approach, the idea with defining a function $F$ is very similar. But to prove that semi-simple rings are Noetherian requires some stuff with composition series, and might rely on the Schreier Refinement Theorem or something like that. I want to avoid using those if possible. Thanks though! $\endgroup$ – Ehsaan May 22 '13 at 21:17

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.