Let $L/K$ be field extension and let $V$ be a $K$-vector space. Then do we have an isomorphism $$L\otimes_K \text {End}_K(V)\cong \text{End}_L(L\otimes_K V)$$ as $L$-algebras?

My attempt: for $\lambda \in L$ and $f\in \text{End}_K(V)$ let $\varphi_{\lambda, f}:L\otimes_K V \to L\otimes_K V$ be the $L$-linear map given by $\alpha \otimes v \mapsto \lambda\alpha\otimes f(v)$.

Then the mapping $\varphi:L\times \text{End}_K(V)\to \text{End}_L(L\otimes V)$ given by $\varphi(\alpha, f)=\varphi_{\alpha,f}$ is $K$-bilinear, so it gives rise to a unique $K$-linear map $\bar{\varphi}:L\otimes_K \text {End}_K(V)\cong \text{End}_L(L\otimes_K V)$ such that $\bar{\varphi}(\lambda\otimes f)=\varphi_{\lambda,f}$. It is also a homomorphism of $L$-algebras.

My problem is showing that $\bar{\varphi}$ has an inverse.

Many thanks.


2 Answers 2


In general Hom and tensor commute when the left argument inside the Hom is finitely presented and the module you're tensoring aganist is flat.

Suppose $V,W$ are modules over a ring $K$ and we're extending scalars through a morphism $K\to L$. Consider the functors $F(V)=L\otimes_K \hom_K(V,W)$ and $G(V)=\hom_L(L\otimes_K V,L\otimes_K W)$. Then we have for each $W$ a map $\eta_V:F(V)\to G(V)$ that sends $1\otimes f$ to the morphism $1\otimes f:L\otimes V\to K\otimes W$ which we extend $L$-linearly. You should check this is an isomorphism if $V$ is free finite. This follows from the natural isomorphisms $\hom_A(A,N)=N$, $A\otimes_A N =N$, the distributivity of tensor over sums, of homs over sums, &c. You just need to see that $\eta_V$ factors as a product of all this known guys. In the general case that $V$ is any $K$-module, suppose you have an exact sequence $L_1\to L_0\to V\to 0$ where $L_i$ are finite free. By using left exactness of hom and that $L$ is flat, we have a commutative diagram of exact rows

$$\newcommand\twoheaduparrow{\mathrel{\rotatebox{90}{$\twoheadrightarrow$}}} \require{AMScd} \begin{CD} 0 @>>> F(V) @>>> F(L_0) @>>> F(L_1) \\ @. @VVV @VVV @VVV \\ 0 @>>> G(V) @>>> G(L_0) @>>> G(L_1) \end{CD}$$

The first and last two vertical arrows are isomorphisms, so that the second arrow is an isomorphism too. If $V$ is assumed to be finitely generated then you get the morphism is injective since you don't have the last arrow.

Example A very useful example in the above is the following, which might be just completely irrelevant to your post: for $M$ a finitely presented $A$-module and $S\subseteq A$ multiplicative, we know $S^{-1}A$ is always $A$-flat so we have that $$S^{-1}\hom_A(M,N)=\hom_{S^{-1}A}(S^{-1}M,S^{-1}N)$$

This can be used to show flat finitely presented $A$-modules are projective.

  • $\begingroup$ So I need to assume that $V$ is finite dimensional? I was wondering if there was an elementary proof of the isomorphism. $\endgroup$
    – mathmo
    Feb 17, 2015 at 17:32
  • 1
    $\begingroup$ You have to use the exactness of the involved functors and the four lemma, using a presentation of the first argument. I can provide the details, or you can check them say in Eisenbud's book on commutative algebra or Bourbaki's book on the same subject. $\endgroup$
    – Pedro
    Feb 17, 2015 at 17:40

It is a general fact of life that there is an isomorphism of $L$-vector spaces: $$\hom_L(L \otimes_K X, Y) \cong \hom_K(X, Y),$$ where $X$ is a $K$-vector space and $Y$ is an $L$-vector space. To see this, if $f \in \hom_K(X, Y)$, let $\bar{f} \in \hom_L(L \otimes_K X, Y)$ be defined by $\bar{f}(\lambda \otimes x) = \lambda f(x)$. The inverse morphism maps $g \in \hom_L(L \otimes_K X, Y)$ to $\tilde{g} \in \hom_K(X,Y)$, where $\tilde{g}(x) = g(1 \otimes x)$.

It follows that: $$\operatorname{End}_L(L \otimes_K V) = \hom_L(L \otimes_K V, L \otimes_K V) \cong \hom_K(V, L \otimes_K V).$$ You want to show that this is isomorphic to $L \otimes_K \hom_K(V,V)$ $(= L \otimes_K \operatorname{End}_K(V))$.

In general there is a canonical linear map: $$\Psi : Z \otimes_K \hom_K(X, Y) \to \hom_K(X, Z \otimes_K Y)$$ that maps $z \otimes f$ ($z \in Z$, $f : X \to Y$) to $\psi_{z,f} : X \to Z \otimes_K Y$, where $\psi_{z,f}(x) = z \otimes f(x)$ (this corresponds to your $\varphi$.) It's also easy to see that when $Z$ is a $K$-algebra (eg. $L$) then this is a morphism of algebras).

Case 1: $L$ is finite dimensional

When $Z$ is finite dimensional over $K$, this map is an isomorphism. Indeed let $(z_k)_{k=1}^n$ be a basis of $Z$ in that case. For $f \in \hom_K(X, Z \otimes_K Y)$, every $f(x)$ can be written as a finite sum $\sum_{k=1}^n z_k \otimes f'_k(x)$, where $f'_k(x) \in Y$. Using the linearity of $f$, you can check that in fact, all the $f'_k$ are $K$-linear maps $X \to Y$, and so you have a well-defined element $$f' := \sum_{k = 1}^n z_k \otimes f'_k \in Z \otimes_K \hom_K(X, Y).$$

Then $f \mapsto f'$ is the inverse of the map $\Psi$ we wrote earlier, which is thus an isomorphism. So in the special case $X = Y = V$, $Z = L$, you get the isomorphism you want (and as I said it is an isomorphism of $K$-algebras), in the case that $L$ is finite-dimensional over $K$.

Case 2: $V$ is finite dimensional

When $X$ is finite-dimensional with basis $(x_k)_{k=1}^n$, $\Psi$ is also an isomorphism. The inverse is even more straightforward: if $f \in \hom_K(X, Z \otimes_K Y)$, let $$f(x_k) = \sum_{i = 1}^{m_k} z_i \otimes y_i.$$

For $y \in Y$ and $1 \le j \le n$, let $\delta_{j,y} : X \to Y$ be the map defined by $$\delta_{j,y}(x_i) = \begin{cases} y & i = j \\ 0 & i \neq j \end{cases}$$

And let $$\theta(f) = \sum_{k=1}^n \sum_{i=1}^{m_k} z_i \otimes \delta_{k, y_i}.$$

Then $\theta$ is the inverse of $\Psi$, so again you get an isomorphism (in the special case $X = Y = V$, $Z = L$) when $V$ is finite dimensional. (This proof essentially comes from the fact $\hom(X,Y) \cong X^* \otimes Y$ when $X$ is finite-dimensional, or "equivalently" $\hom(K^n, Y) \cong Y \oplus \dots \oplus Y$ and $Z \otimes (U \oplus V) \cong Z \otimes U \oplus Z \otimes V$).

  • 1
    $\begingroup$ When $L$ isn't finite-dimensional things get more complicated. It seems the dimension of $V$ starts playing a role. Unfortunately what precedes is too long for a comment and I'm not sure I will have the time to finish writing down the infinite dimensional case today. $\endgroup$ Feb 17, 2015 at 18:14

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