# Existence of tensor product of subalgebras

Let $\mathcal{G} = \mathbb{M}_n(\mathbb{C})$ be an $n$-by-$n$ matrix algebra over complex numbers. Next let $\mathcal{A} \cong \mathbb{M}_d(\mathbb{C})$ be a subalgebra of $\mathcal{G}$ and assume $d$ divides $n$. Then is it true that there exists another subalgebra of $\mathcal{G}$ (let's call it $\mathcal{B}$) such that

1) $\mathcal{G} \cong \mathcal{A} \otimes \mathcal{B}$.

2) Every element in $\mathcal{A}$ commutes with every element in $\mathcal{B}$.

3) The intersection of $\mathcal{A}$ and $\mathcal{B}$ are multiples of the identity.

If it is not true, what are the necessary conditions?

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## 3 Answers

Yes, those are true and are consequences of the Double Centralizer Theorem for central simple algebras. See, for example, section 12.7 in Pierce's Associative Algebras.

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Mariano has answered it but I'd like to add that $d$ divides $n$ is a red herring. If $M_d (C)$ is a subalgebra of $M_n(C)$ then decomposing $C^n$ into a direct sum of simple $M_d(C)$ modules tells you that $d$ must divide $n$.

This also tells you that the centraliser of $M_d(C)$ is isomorphic to $M_{n/d}(C)$ and all your conditions 1,2,3 follow trivially...

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so you are saying the fact that $\mathcal{A} \cong \mathbb{M}_d$ is a subalgebra of $\mathcal{G}$ implies $d$ must divide $n$? –  kett Nov 11 '10 at 2:19
If $\mathcal{A}$ has the same unit as $\mathcal{G}$, then yes. –  Martin Argerami Nov 13 '10 at 2:59
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Just to contribute a little, let me give a von Neumann algebra perspective.

Let $B=A'$, i.e. the commutant of $A$ in $G$. Then 2) is immediately satisfied, and so is 3), because $A\cap B$ is the centre of $A$, which is trivial.

As we are in a type I factor, $B'=A$.

Also $$\text{alg}(A,B)=(A\cup B)''=(A'\cap B')'=(A'\cap A)'=(\mathbb{C}\,I)'=G,$$ which shows that $G$ is the algebra generated by $A$ and $B$. It remains to see that $G$ can be identified with $A\otimes B$. The map $a\times b\mapsto ab$, extended by linearity, is a *-homomorphism from $A\otimes B$ to $G$. It is clearly onto. And it is also one-to-one, because if $\sum_ja_jb_j=0$, one can use the idea in 11.1.8 in Kadison-Ringrose to see that $\sum_ja_j\otimes b_j=0$ (it is essential the fact that $a_jb_j=b_ja_j$). Finally, a bijective *-homomorphism is also isometric.

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