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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.

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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 $\{k_1,..,k_l\}=\{i_1,..,i_p\}\cap\{j_1,..,j_q\}$.

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  • $\begingroup$ $I$ 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. $\endgroup$ – Aaron Zolotor Oct 26 '18 at 18:46
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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

$$x=xj \in IJ$$

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

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