# Tensor products over field do not commute with inverse limits?

In the question: Inverse limit of modules and tensor product, Matt E gives an example where inverse limits and tensor products do not commute over the base ring $\mathbb{Z}$. He then goes on to show that it does hold if one takes a limit over modules of finite length and tensors with a finitely presented module. Are there counterexamples in the category of vector spaces over a field $k$ (not necessarily finite dimensional of course)?

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Yes. Take a field $k$. Then $k[[t]]$ is (isomorphic to) the product of infinitely many copies of $k$. If $V = k^\infty$ is an infinite dimensional space, then the canonical map $k[[t]] \otimes_k V \to V[[t]] \simeq \prod_{n=1}^\infty k \otimes V$is not an isomorphism: its image is the subspace of elements of the product such that the coefficients span a finite dimensional subspace of $V$.

Indeed, elements of its image are of the form $x(t) = \sum_{k=1}^d R_k(t) \otimes v_k$ where $R_k$ are power series and $v_k \in V$, and the coefficients of $x$ are in the vector space $\mathrm{Span}(v_1, \dots, v_d)$ which is finite dimensional. On the other hand, if $(e_i)$ is a basis of $k^\infty$, then $\sum_{i=1}^\infty e_i t^i$ is not in the image.

This example is the motivation for the definition of the completed tensor product, by the way.