Is $M_2(\Bbb R)\otimes_{\Bbb R}\Bbb C\cong M_2(\Bbb C)$?

Question

Is $M_2(\Bbb R)\otimes_{\Bbb R}\Bbb C\cong M_2(\Bbb C)$? What about more generally, is $M_n(k)\otimes_k K\cong M_n(K)$ for $K/k$ algebraic?

I think $M_2(\Bbb R)\otimes_{\Bbb R}\Bbb C\to M_2(\Bbb C)$ sending $e_{ij}\otimes 1\mapsto e_{ij}$ is an isomorphism of $\Bbb C$-modules. We only have to define the map on these four generators, and then extend $\Bbb C$-linearly I believe. In which case $z\cdot (e_{ij}\otimes 1)=(e_{ij}\otimes 1)\mapsto ze_{ij}$. Then this is surjective since $$\begin{bmatrix}a&b\\c&d\end{bmatrix},\qquad a,b,c,d\in\Bbb C$$ is mapped to by $e_{11}\otimes a+e_{12}\otimes b+e_{21}\otimes c+e_{22}\otimes d$ and it is injective since the kernel is trivial. For the multiplicative part of the $\Bbb C$-algebra homomorphism we just have to check pair-wise products I think?

$$\varphi((e_{ij}\otimes 1)(e_{kl}\otimes 1))=\delta_{jk}\varphi(e_{il}\otimes 1)=\delta_{jk}e_{il}$$ $$\varphi(e_{ij}\otimes 1)\varphi(e_{kl}\otimes 1)=e_{ij}e_{kl}=\delta_{jk}e_{il}$$ is this all correct? What about that $M_n(k)\otimes_k K\cong M_n(K)$?

Answer 1

If $K/k$ is a field extension, define $$ K \times M_n(k) \ni (\lambda, A) \mapsto \lambda A \in M_n(K). $$This is $k$-balanced, so it induces a map $K\otimes_k M_n(k) \to M_n(K)$. If $E_{ij} \in M_n(K)$ is the matrix with $1$ in position $(i,j)$ and $0$ elsewhere, every matrix $A \in M_n(K)$ writes uniquely as $A = \sum_{i,j}a_{ij}E_{ij}$, where $a_{ij} \in K$. So put $$M_n(K) \ni \sum_{i,j}a_{ij}E_{ij} \mapsto \sum_{i,j} a_{ij}\otimes E_{ij} \in K\otimes_kM_n(k). $$This is the inverse map. Thus $K\otimes_k M_n(k) \cong M_n(K)$.

Answer 2

We have in general for a field $k$ and an extension $K$ of it: $$ k^n\otimes_k K=K^n\ . $$ Here $n=1,2,3,\dots$ is a positive integer. The OP is a special case with $n=2^2$.

(The multiplication operations of matrices correspond via this isomorphism, denoted above even with the equality sign.)

We use $(V\oplus W)\otimes _k K = (V\otimes_k K)\; \oplus\; (W\otimes_k K)$.

To have a more general setting, see also

Tor functor .

In the category of vector spaces over a field, each module = vector space is free, thus flat. For more general cases, $R$ algebra over a ring $S$, start with a short exact sequence $0\to M'\to M\to M''\to 0$ instead of (the split short exact sequence) $0\to V\to V\oplus W\to W\to W\to 0$, then apply the functor $M\to M\otimes _S R$ to get a long exact sequence, then ask for the vanishing of the corresponding Tor-object.