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While preparing for a test I found the next question which i cannot fully answer:

Assume $k$ is an algebraically closed field, and $g_{1},g_{2}$ are $k$-Lie algebras and let $g=g_{1}\times g_{2}$. The first part of the question asks: If $V_{1},\, V_{2}$ are irreducible $g_{1},\, g_{2}$ modules (respectively) then $V=V_{1}\otimes V_{2}$ is an irreducible $g$-module. That I could do.

The second part however asks: Show that any ireducible g-module is isomorphic to some $V_{1}\otimes V_{2}$ as above. This I don't know how to prove.

There is also a question in the end which asks what is the difference if $k$ is not algebraically closed.


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I would say that given a $g_1 \times g_2$ module $V$, you should consider it as a $g_1$ module, take its irreducible submodule $V_1$, and consider the space $Hom_{g_1}(V_1, V)$ as a $g_2$ module. Then the tensor product over $k$ of $V_1$ with the Hom space will be a module over $g_1 \times g_2$ and a submodule of $V$, thus equal to $V$. You then show at the Hom is an irreducible representation of $g_2$.

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If $V_2={\sf Hom}_{g_1}(V_1,V)$ and $W=V_1 \otimes V_2$, we have a natural map $W \to V, v_1 \otimes \phi \mapsto \phi(v_1) $. This map is not an embedding, however (it is clearly not injective in general), so you cannot view $W$ as a submodule of $V$. – Ewan Delanoy Apr 5 '12 at 8:40

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