Let $K$ be an algebraically closed field, and let $A$ and $B$ be semisimple finite-dimensional $K$-algebras. I've seen a claim that the tensor product $A \otimes_K B$ is also a semisimple ring.

To prove this claim, I've thought of using the Wedderburn-Artin theorem, which tells us that $A \cong \prod_{i = 1}^m M_{n_i}(K)$ and $B \cong \prod_{j = 1}^{m'} M_{n_j'}(K)$ for some natural numbers $n_i$ and $n_j'$. It can be shown that tensor products preserve finite direct products of rings, and hence $$A \otimes_K B \cong \prod_{i, j} M_{n_i}(K) \otimes_K M_{n_j'}(K).$$ It therefore suffices to show that $A \otimes_K B$ is semisimple in the special case where $A$ and $B$ are matrix rings over $K$. But how can we prove this?

  • 2
    $\begingroup$ $ M_{n_i}(K) \otimes_K M_{n_j'}(K)\cong M_{n_i n_j}(K)$ which is simple $\endgroup$
    – reuns
    Mar 8, 2022 at 13:13

1 Answer 1


As reuns mentioned in a comment, the tensor product of two matrix rings is still a matrix ring, which is semisimple. Since I hadn't seen this result before, I've decided to provide a proof.

It suffices to show that if $V$ and $W$ are finite-dimensional vector spaces over $K$, then $\operatorname{End}_K(V) \otimes_K \operatorname{End}_K(W) \cong \operatorname{End}_K(V \otimes_K W)$ as rings. By the universal property of the tensor product, there is a $K$-linear map \begin{align} \phi \colon \operatorname{End}_K(V) \otimes_K \operatorname{End}_K(W) &\to \operatorname{End}_K(V \otimes_K W) \\ f \otimes_1 g &\mapsto f \otimes_2 g, \end{align} where $f \otimes_1 g$ denotes the elementary tensor of $f$ and $g$, and $f \otimes_2 g$ denotes the unique endomorphism of $V \otimes_K W$ that maps a tensor of the form $v \otimes w$ to $f(v) \otimes g(w)$. It is straightforward to check that $\phi$ is also a ring homomorphism.

We need to check that $\phi$ is a bijection. Since $\phi$ is a linear map between finite-dimensional vector spaces of equal dimension, it is enough to check that $\phi$ is surjective. We will do this by using Kronecker products. Let $\{v_1, \ldots, v_m\}$ and $\{w_1, \ldots, w_n\}$ be ordered bases for $V$ and $W$, respectively. Let $f\colon V \to V$ and $g \colon W \to W$ be endomorphisms, and let $F$ and $G$ be the matrices for $f$ and $g$, respectively, with respect to the ordered bases. Then it is fairly straightforward to show that with respect to the ordered basis $$\{v_1 \otimes w_1, v_1 \otimes w_2, \ldots, v_1 \otimes w_n, \ldots, v_m \otimes w_1, v_m \otimes w_2, \ldots, v_m \otimes w_n \},$$ the matrix for $f \otimes_2 g$ is the Kronecker product $F \otimes G$, which is given by the following block matrix: $$ F \otimes G = \begin{bmatrix} f_{11}G & \cdots & f_{1m}G \\ \vdots & \ddots & \vdots \\ f_{m1}G & \cdots & f_{mm}G \end{bmatrix} $$

In order to prove that $\phi$ is surjective, we therefore only need to show that every matrix in $M_{mn}(K)$ can be written as a finite sum of Kronecker products. So let $C$ be an $mn\times mn$ matrix. We can write $C$ in block matrix form as $$ C = \begin{bmatrix} C_{11} & \cdots & C_{1m} \\ \vdots & \ddots & \vdots \\ C_{m1} & \cdots & C_{mm} \end{bmatrix}, $$ where $C_{ij}$ is an $n\times n$ matrix. Let $G_1, \ldots, G_t$ be $n \times n$ matrices that span $M_n(K)$. Then for every $C_{ij}$, there exist scalars $f_{ij}^k$ such that $$C_{ij} = \sum_{k = 1}^t f_{ij}^k G_{k}.$$ Then $C$ can be written as a sum of Kronecker products: \begin{align} C &= \sum_{k} \begin{bmatrix} f_{11}^k G_k & \cdots & f_{1m}^k G_k \\ \vdots & \ddots & \vdots \\ f_{m1}^k G_k & \cdots & f_{mm}^k G_k \end{bmatrix} \\ &= \sum_k F_k \otimes G_k, \end{align} where $F_k = (f_{ij}^k)$. This shows that $\phi$ is surjective, as desired.


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