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Corollary of Artin Wedderburn: (p18, Introudction to finite group theory, Webb) Let $A$ be a finite dimensional semsimple algebra over an algebrically closed field $k$. In any decomposition $$ _AA = S_1^{n_1} \oplus \cdots \oplus S_r^{n_r} \cong M_{n_1}(D_1) \oplus \cdots \oplus M_{n_r}(D_r)$$ where $S_i$ are pairwise nonisomoprhic simple modules, and $D_i=End_A(S_i)^{op}$. We have $n_i = \dim_k S_i$.

For the proof the author begins with

$M_{n_i}(k) \cong S_i^{n_i}$ as left $A$-modules, since term on left is isomorphic to the quotient of $A$ by the left submodules consisting of elements that the summand $M_{n_i}(k)$ annihilates by right multiplication, and the term on right is an image of this quotient and we have $\dim_k(A) = \sum_i \dim_k S_i^{n_i}$ they must be isomorphic.

I am really confused, what are the explicit maps involved here?

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Why you don't just act $A=\bigoplus_{i=1}^rS_i^{n_i}$ by $\mathrm{End}_A$ for this corollary? If so, one has $$ \mathrm{End}_A(A)=\bigoplus_{i=1}^r\mathrm{End}_A(S_i^{n_i})\cong\bigoplus_{i=1}^rM_{n_i}(D_i^{\mathrm{op}}). $$ Moreover, $A^\mathrm{op}\cong\mathrm{End}_A( _AA)$ is given by $a\mapsto(x\mapsto xa)$, and hence $$ A^\mathrm{op}\cong\bigoplus_{i=1}^rM_{n_i}(D_i^{\mathrm{op}})\cong\bigoplus_{i=1}^rM_{n_i}(D_i)^{\mathrm{op}}, $$ which implies $$ A\cong\bigoplus_{i=1}^rM_{n_i}(D_i). $$

I don't understand why in your corollary $A$ is assumed to be over an algebraically closed field $k$, since in this case, $D_i$'s are finitely dimensional division algebras over $k$ and hence all equal to $k$.


In allusion to your question, as is said above, $k\cong D_i^\mathrm{op}\cong \mathrm{End}_A(S_i)$, and hence $M_{n_i}(k)\cong\mathrm{End}_k(S_i)$. Then the Jacobson density theorem provides an epimorphism \begin{align} \varphi\colon A&\twoheadrightarrow \mathrm{End}_k(S_i)\\ a&\mapsto(x\mapsto xa). \end{align} Pick a $k$-basis $\{e_1,\cdots,e_{n_i}\}$ of $S_i$. Define \begin{align} \Phi\colon\mathrm{End}_k(S_i)&\twoheadrightarrow S_i^{n_i}\\ f&\mapsto (f(e_1),\cdots,f(e_{n_i})) \end{align} and $\Phi$ is also epimorphic. These are the explicit maps in your questioned part.

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