# Tensor product of finite-dimensional semisimple algebras over algebraically closed field is semisimple

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?

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

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.