How to prove that finite dimensional von Neumann algebra is direct sum of matrix algebras?
deleted by user21820, Xander Henderson, Saad Apr 5 at 17:34
closed as offtopic by José Carlos Santos, user99914, Mostafa Ayaz, Stefan4024, rschwieb Aug 2 '18 at 23:03
This question appears to be offtopic. The users who voted to close gave this specific reason:
 "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Community, Mostafa Ayaz, Stefan4024, rschwieb

$\begingroup$ All right ideals are finitely generated, hence summands. QED. But next time, please don’t dump your questions without some context. For example, maybe you expect to reprice the entire artinian wedderburn theorem: I don’t know, because you didn’t say. $\endgroup$ – rschwieb Aug 2 '18 at 23:05
You first show that there is a finite (unique) set $p_1,\ldots,p_m$ of minimal central projections.
Then you prove that $p_jMp_j=p_jM$ is a factor. Since $M=\bigoplus_j p_jM$, all that remains is to show that a finitedimensional factor is isomorphic to $M_n(\mathbb C)$ for some $n$. You start with a minimal projection $e_{11}$, and construct pairwise orthogonal minimal projections $e_{11},\ldots,e_{nn}$. Since we are now in a factor, minimal projections are equivalent. So there exist partial isometries $e_{12},\ldots,e_{1n}$ such that $$ e_{11}=e_{1j}^*e_{1j},\ \ \ e_{jj}=e_{1j}e_{1j}^*. $$ With these partial isometries we define $$ e_{kj}=e_{1k}^*e_{1j}. $$ And then one can check that these behave exactly as the matrix units in $M_n(\mathbb C)$, namely $$ e_{kj}e_{st}=\delta_{js}\,e_{kt}, $$ and $e_{11}+\cdots+e_{nn}=I_n$.
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