Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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 1

up vote 2 down vote accepted

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.