# $R$ semisimple $\implies IJ = I \cap J$

Let $$R$$ be a semisimple ring, $$I$$ be a right ideal of $$R$$ and $$J$$ be a left ideal. We wish to show $$IJ = I \cap J$$.

Clearly $$IJ \subseteq I \cap J$$. Since $$R$$ is semisimple and $$J$$ is a left $$R$$-modules and a submodule of the left regular module we have

$$R=J \oplus L$$

for some complement $$L \subseteq R$$ of $$J$$. Further, $$IJ$$ and $$I \cap J$$ are both submodules (in fact, they are ideals) so we have

$$R = I \cap J \oplus K \quad \text{and} \quad R=IJ \oplus S$$

Further, $$I \cap J$$ is semisimple and $$IJ$$ is a submodule of $$I \cap J$$ so we have

$$I \cap J= IJ \oplus Q$$

I've tried various ways of playing with the above equalities and I believe that I need to find a complement in a clever way such that it shows the $$Q$$ above must be 0. But I am quite stuck.

## 2 Answers

If $$R$$ is semi-simple, $$R=I_1+...+I_n$$ where $$I_j$$ is simple, $$I=I_{i_1}\oplus..\oplus I_{i_p}$$, $$J=I_{j_1}\oplus..\oplus I_{j_q}$$ where $$\{i_1,..,i_p\}$$ and $$\{j_1,...,j_q\}$$ are subsets of $$\{1,...,n\}$$.

$$I_iI_j=0$$ if $$i\neq j$$ and $$I_iI_i=I_i$$, This shows that $$IJ=(I_{i_1}\oplus..\oplus I_{i_p})(I_{j_1}\oplus..\oplus I_{j_q})=I_{k_1}\oplus..\oplus I_{k_l}=I\cap J$$ where $$\{k_1,..,k_l\}=\{i_1,..,i_p\}\cap\{j_1,..,j_q\}$$.

• $I$ is a right ideal. Hence, it isn't a submodule of $R$. I don't think you can write it as a direct sum of simple submodules. – Aaron Zolotor Oct 26 '18 at 18:46

It is clear the $$IJ \subseteq I \cap J$$. Now, we have

$$R = J \oplus L$$ since $$R$$ is a minimal ideal. Hence, for some $$j \in J$$ and $$l \in L$$ we have

$$1=j+l$$.

Now, let $$x \in I \cap J$$ and conisder $$x=xj+xl$$. Then, in particular, $$x \in J$$ so $$xl=0$$. Hence

$$x=xj \in IJ$$

• $R$ is a minimal ideal? I doubt this is what you wanted to say here :) – darij grinberg May 30 at 21:59