$M_n(k)\otimes_kB\cong M_n(B)$, where $k$ is a field, and $B$ is a $k$-algebra.

Question

Let $k$ be a field and let $B$ be a $k$-algebra, then $M_n(k)\otimes_kB\cong M_n(B)$.

There is no need to read the following texts.

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A brief overview of what I know, and what I did:

Let $E_{ij}$ be the matrice with 1 in the entry $ij^{\rm th}$, and $0$ elsewhere. Then $E_{ij}$'s form a basis for $M_n(k)$. $M_n(k)\otimes_kB$ is a free $B$-module, with the generating set $\{E_{ij}\otimes1_B\}$. Also we have these relations: $(E_{ij}\otimes1_B)(E_{kl}\otimes1_B)=\delta_k^j(E_{il}\otimes1_B)$. I can not go further, and even I can not recognize if I am walking in a suitable direction or not.

I solved some exercises on tensor products, most of them are solved with the same idea for this problem: Let $G$ be an abelian group, such that the order of any element is finite. Then $G\otimes_{\mathbb{Z}}\mathbb{Q}\cong0$. But that problem has a different nature, and I stuck in that problem, and I don't know how should I show that.

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Final Edit: (My solution to $M_n(A)\otimes_kB\cong M_n(A\otimes B)$, where $k$ is a field, and $A$ and $B$ is a $k$-algebras.) (I am not sure if my solution is true or not)

$M_n(A)\otimes_kB$ is a free $B$-module, with the basis $\{(E_{ij}\otimes1_B) \mid 1 \leq i, j \leq n \}$. Now define the map $\varphi$ on the elements of basis as follows:

\begin{gather*} \varphi:M_n(A) \otimes_k B \longrightarrow M_n(A\otimes B)\\ (E_{ij}\otimes1_B) \mapsto E_{ij} \end{gather*}

Then we have

$$\varphi((E_{ij}\otimes1_B)(E_{kl}\otimes1_B))=\varphi(E_{ij}E_{kl}\otimes 1_B)=\varphi(\delta_k^jE_{il}\otimes 1_B)=\delta_k^jE_{il}=E_{ij}E_{kl}=\varphi(E_{ij}\otimes1_B)\varphi(E_{kl}\otimes1_B)$$

Answer 1

As $k$-modules, $M_n(k)$ is isomorphic to $k^{(n^2)}$, and similarly, $M_n(B) \sim B^{(n^2)}$. Therefore, since tensor product distributes over direct sum, $M_n(k) \otimes_k B \sim M_n(B)$ as $k$-modules. The isomorphism is given via $(x_1, ... x_{n^2}) \otimes b \rightarrow (b x_1 , ... b x_{n^2})$. Call this function $\phi$.

The only thing we need to do now is to show that this preserves multiplication. By the distributive property, it suffices to show that it preserves multiplication on pure tensors.

Let $A, B \in M_n(k), x, y \in B$. Then, $\phi(A \otimes x * B \otimes y) = \phi(AB \otimes xy) = xy AB$, and $\phi(A \otimes x) * \phi(B \otimes y) = xA * yB = xy AB$.

Therefore, it preserves multiplication and is an isomorphism.

Answer 2

EDIT: as pointed out in the comments the approch is wrong but I still think it is a bit valuable so I'm not deleting it for now.

I'll try. $B$ is a $k$-algebra hence there exists a ring homorphism $T:k\rightarrow B$ such that $T(k)$ is in the center of $B$. Given a matrix $M \in M_n(k)$ let me extend the notation naming $T(M) \in M_n(B)$ the matrix obtained applying $T$ elementwise.

Consider the following map \begin{gather*} \phi:M_n(k) \otimes_kB \longrightarrow M_n(B)\\ \phi(M \otimes b) = b \ T(M) \end{gather*}

The map such defined is a $k$-module homomorphism. Considering that the elements $E_{ij}\otimes b$ are mapped into the elements $bE_{ij}$ of $M_n(B)$ we can conclude that the map is surjective.

Now suppose two elements have the same image, so

\begin{gather*} \phi(M_1 \otimes b_1) = \phi(M_2 \otimes b_2) \\ b_1 T(M_1) = b_2 T(M_2) \end{gather*}

This means

\begin{gather*} \forall(i,j) \ \ m_{ij}^1 b_1 = m_{ij}^2 b_2 \\ \forall(i,j) \ \ b_2 = (m_{ij}^2)^{-1}m^1_{ij} b_1 \\ \end{gather*}

So for each $(i,j)$ the ratio $r=(m_{ij}^2)^{-1}m^1_{ij}$ is constant, and we have $b_2 = rb_1$, so

\begin{gather*} M_2 \otimes b_2 = M_2 \otimes rb_1 = rM_2 \otimes b_1 = M_1 \otimes b_1 \end{gather*}

demonstrating that $\phi$ is an injection on simple tensor.