Does tensoring with a finite-dimensional module preserve products? Let $R$ be any $k$-algebra, $k$ a field. Suppose $M$ is a finite dimensional $R$-module. When does $M \otimes_{R} - $ commutes with arbitrary direct products? Any hints on why this is true? Is there any condition on $R$ that will make this true?
 A: 
Theorem: For any ring $R$ and any right $R$-module $M$, the functor $M\otimes_ R - : R\textsf{-Mod}\to\textsf{Ab}$ commutes with products if and only if $M$ is finitely presented. 

This is Proposition 4.44 in Lam, Lectures on Modules and Rings. 
In your case, $M$ is finite-dimensional over $k$, so in particular finitely generated over $R$, but not necessarily finitely presented without further assumptions on $R$.
Example: Take $R = k[x_1,x_2,\ldots]$ and $M = R/(x_i) = k$, which is finitely generated but not finitely presented. Therefore, according to the Theorem, $k\otimes_R - $ shouldn't commute with arbitrary products, and indeed: For any $R$-module $N$ we have $k\otimes_R N \cong N / (x_i) N$ naturally, so putting $N_n := k[x_n^{\pm 1}]$, with all $x_i, i\neq n$ acting as zero, we have $k\otimes_R N_n=0$ for all $n$, but $k\otimes_R \prod_n N_n\neq 0$ since $$(x_i)N=(x_i)\prod_n N_n\subset\bigoplus_n N_n\subsetneqq\prod_n N_n=N.$$
If, however, you impose finite-dimensionality of $R$, then $R$ is Noetherian and hence $M$ being finite-dimensional over $k$ implies it being finitely presented over $R$, and the statement holds.
