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?

  • $\begingroup$ 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

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.)


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