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I want to prove that $Mat_{n_1}(k) \otimes_k Mat_{n_2}(k) \cong Mat_{n_1n_2}(k) $ (as $k$-algebras) where $k$ is a field by checking the universal property.

Namely, I need to inclusion $Mat_{n_{1,2}}(k) \hookrightarrow Mat_{n_1n_2}(k)$ -- that must be obvious.I can put $n_1 \times n_1$ matrix to the up-left and $n_2 \times n_2$ to the down-right.

Now I need to show that for any $k$-algebra $A$ and for any $\phi_i: Mat_{n_i}(k) \to A $, such that $\phi_1(A)$ commures with $\phi_2(B)$ for any $A \in Mat_{n_{1}}(k), \; B \in Mat_{n_{2}}(k) $ there is a unique $\Phi: Mat_{n_1n_2}(k) \to A$ such that $\Phi \circ \mathcal{i}_i = \phi_i. $

In other words, the image of $\Phi: Mat_{n_1n_2}(k) \to A$ must be fully determined by what it does to those two subalgebras. I think that it is actually determined by what it does to diagonal elements but I couldn't prove it.

I know that diagonal elements form a basis for matrix algebra as module over itself but it doesn't help. Also my guess never uses that images of metrices from different subalgebras commute.

Invincible
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I think you can proceed in the following way: The map

$$Mat_{n_1}(k) \times Mat_{n_2}(k) \to Mat_{n_1n_2}(k): (A,B) \mapsto A \odot B$$

is $k$-bilinear. Here $A \odot B$ denotes the kronecker-product of matrices.

Consequently, the universal property of the tensor product gives a $k$-algebra morphism

$$f: Mat_{n_1}(k)\otimes Mat_{n_2}(k) \to Mat_{n_1n_2}(k)$$

where $f(A \otimes B) = A \odot B$.

Show that $f$ is an isomorphism of $k$-algebra's.

J. De Ro
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  • But it gives a $k$-module morphism, not $k$-algebra. – Invincible Apr 09 '20 at 13:46
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    Then exercise for you it preserves the ring structure! – J. De Ro Apr 09 '20 at 13:48
  • I know that the Kronecker product of matrices is not the same as the tensor product of matrices (in case such a thing really exists according to this answer: https://math.stackexchange.com/questions/203947/tensor-product-and-kronecker-product) but then...what is $A\otimes B$ in the answer? Can it be done the same replacing $A\odot B$ by $A\cdot B$, ie: the usual product of matrices? – gibarian Oct 19 '21 at 06:42
  • @gibarian The usual matrix product has dimension $n_1+n_2$ while we need dimension $n_1n_2$. – J. De Ro Oct 19 '21 at 07:43
  • In this answer $A\otimes B$ is the abstract tensor product of vector spaces. – J. De Ro Oct 19 '21 at 07:44