6
$\begingroup$

It is pretty routine to show that every simple left $R$-module is cyclically generated by any nonzero element and that these simple modules are precisely of the form $R/I$, where $I$ is some maximal ideal.

My question is what happens when you start to 'stitch' these together: let $S_1,S_2$ are distinct (in that they are not isomorphic) simple left $R$-modules. Then these are of the form $R/I, R/J$, where $I,J$ are maximal left ideals, respectively. My thought it is that $S_1 \oplus S_2 \cong R/(I \cap J)$. This makes since in that if $R$ were semisimple, then it is a sum of simple modules and its jacobson radical (the intersection of all maximal left ideals) is zero. Then extending my idea above, $I \cap J$ does seem to be the correct ideal. But I am unable to prove this.

My idea was to define a surjective map $\phi: R \to R/I \times R/J$ and use the first isomorphism theorem but the only obvious map is $r \mapsto (r,r)$ and its not clear that is surjective. [For a bit I thought it wouldn't be but perhaps it, non-obviously is?] I could go the other way, $\psi: R/I \times R/J \to R/(I \cap J)$ via $(r,s) \mapsto rs+I\cap J$ (easy enough to show this is well-defined since this vanishes on $I \cap J$). This map is pretty clearly surjective. But injective? If $(r,s)$ were to map to zero, then $rs \in I \cap J$. But beyond that I'm not sure where to go with this. If the ring were commutative, I would have more since maximal implies prime and this would be something. (since this doesn't work, I'm really out of ideas)

Any ideas on how to prove this or can $S_1 \oplus S_2$ not be represented this way?

$\endgroup$
  • $\begingroup$ You should say “left” (or right) maximal ideal. It makes a big difference: a simple left module has the form $S/I$, where $I$ is a maximal left ideal. The Jacobson radical is the intersection of the maximal left ideals, in general distinct from the intersection of the maximal two-sided ideals. What do you mean by “distinct”? Is it in the sense of “not isomorphic”? $\endgroup$ – egreg Jan 9 '17 at 20:48
  • $\begingroup$ @egreg Yes, of course, it was what I was thinking. I shall make the edit. $\endgroup$ – TinyTim Jan 9 '17 at 20:49
  • $\begingroup$ @RobArthan Meaning non-isomorphic. $\endgroup$ – TinyTim Jan 9 '17 at 20:50
  • $\begingroup$ By the way, your map $R/I\times R/J\to R/(I\cap J)$ is not a homomorphism of $R$-modules. $\endgroup$ – egreg Jan 9 '17 at 20:58
2
$\begingroup$

Your diagonal map $\phi:R\to R/I\times R/J$ is indeed surjective, giving an isomorphism $R/(I\cap J)\cong R/I\times R/J$. To show it is surjective, it suffices to show that $(1,0)$ is in its image, since $(1,0)$ and $\phi(1)=(1,1)$ generate $R/I\times R/J$ as an $R$-module.

So we want to find $r\in R$ such that $r\in J$ and $r-1\in I$. Since $I\neq J$ and $J$ is maximal, there exists $s\in J\setminus I$. By maximality of $I$, the left ideal generated by $I\cup \{s\}$ is all of $R$, so we can write $1=as+i$ for some $a\in R$ and $i\in I$. We can then take $r=as$, and we have $r\in J$ since $s\in J$ and $r-1=-i\in I$. We then have $\phi(r)=(1,0)$, as desired.

(Note that this argument only needs that $I\neq J$, not that $S_1\not\cong S_2$. It is possible to have $S_1\cong S_2$ but $I\neq J$; for instance, if $R=M_n(k)$ is a matrix ring over a field, then any two simple $R$-modules are isomorphic.)

$\endgroup$

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.