# R semisimple ring implies R direct sum of finite number of minimal left ideals.

I'm working through a proof of this statement which goes roughly as follows:

"The simple submodules of $R$ (as a module over $R$) are exactly the minimal left ideals of $R$. So (from earlier theorem) $R = \bigoplus_{i \in I} S_i$ where each $S_i$ is a minimal left ideal...''

Then comes the part that I don't understand:

"...In particular the element $1 \in R$ can be written as a finite sum, \begin{eqnarray*} 1 = x_{i_1} + \dots + x_{i_n} \end{eqnarray*} where $x_{i_j} \in S_{i_j}$. It then follows by multiplication of $r = r \cdot 1$...''

I don't see why $1$ can be written as this finite sum?

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Because everything in $R$ can be written as such a sum if $R = \bigoplus_{i \in I} S_i$; look at the definition of direct sum. – Brian M. Scott Jan 18 '12 at 15:32

## 1 Answer

This is not a special property of the element $1$. In fact, every element $x\in R = \bigoplus_{i \in I} S_i$ can be written as a finite sum \begin{eqnarray*} x = x_{i_1} + \dots + x_{i_n} \end{eqnarray*} where $x_{i_j} \in S_{i_j}$ and $i_j\in I$.

This is a direct consequence of the definition of direct sums of rings.

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I’d go ahead and link; it’ll make the answer more useful in the future. – Brian M. Scott Jan 18 '12 at 15:40
Note that quickly pressing "esc" before the wikipedia page loads allows you to remain on the page (though some of the images don't appear) – Daniel Freedman Jan 18 '12 at 15:45
of dear! Of course. Thank you! – user50948 Jan 18 '12 at 16:40