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The Wedderburn-Malcev theorem states that for every (associative unital) finite-dimensional algebra $A$ over field $F$, there exist a semisimple subalgebra $S$, such that $A=\mathrm{Rad}(A)\oplus S$ as a vector space, where $\mathrm{Rad}(A)$ is the Jacobson radical of $A$.

Can someone help me about the structure of subalgebra $S$? I mean I know the existence of of such $S$, is it possible to understand how it is constructed and what is its relation with $A$?

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    $\begingroup$ I think $A/rad(A)$ has to be separable. $\endgroup$
    – Todd Leason
    Commented Mar 18, 2016 at 19:20
  • $\begingroup$ According to this, the part you cite is still due to Wedderburn. Malcev's contribution was to show that any semisimple $S'$ such that $A = Rad(A) \oplus S'$ must be conjugate to $S$ by an element of the form $(1+r)$ for $r \in Rad(A)$. $\endgroup$
    – Joshua Grochow
    Commented Jun 1, 2019 at 4:15

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$S \cong A/rad(A)$ as $F$-algebra.

Proof: The composition $S \hookrightarrow A \twoheadrightarrow A/rad(A)$ is a surjective $F$-algebra homomorphism with kernel $S \cap rad(A)=0$.

By Wedderburn's theorem we also know $A/rad(A) \cong \prod_{i=1}^m M_{n_i}(D_i)$ where $D_i$ is a division algebra over $F$.

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I have also asked this question and partially answered within my diploma thesis (Separabilität in kommutativen und auflösbaren Algebren. Unter Berücksichtigung nicht-unitärer assoziativer Algebren).

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