Minimal and characteristic polynomials on tensor product spaces Given two finite-dimensional vector spaces $V$ and $W$ over a common field $k$ as well as $k$-linear transformations $\varphi \colon V \to V$ and $\psi \colon W \to W$, what can be said in general about the minimal and characteristic polynomials of the tensor product of the linear transformations? That is, can one describe the minimal and characteristic polynomials of $\varphi \otimes \psi \colon V \otimes W \to V \otimes W$ in terms of the minimal and characteristic polynomials of $\varphi$ and $\psi$?
Moreover, what do we glean in addition by letting $\psi$ be the identity matrix? (Edit 1: the case where $\psi$ is the identity matrix for both the characteristic and minimal polynomials is explained by @darij grinberg in the comments.)
Edit 2: @darig grinbeg's comments outline that the characteristic polynomial of $\varphi \otimes \psi$ is indeed determined by the characteristic polynomials of $\varphi$ and $\psi$, though we don't yet have a `nice' relation. Nothing has been said about the minimal polynomial.
 A: This is a compilation of @darij grinberg's comments, which provide a partial answer to the question.
There is an operation that takes a monic polynomial $P$ of degree $n$ and a monic polynomial $Q$ of degree $m$ and returns a monic polynomial $R$ of degree $nm$ whose roots are the products of each root of $P$ with each root of $Q$ (in a suitable extension of $k$). If $P$ is the characteristic polynomial of $\varphi$, and $Q$ is that of $\psi$, then $R$ is that of $\varphi \otimes \psi$. Writing the $\ell$-th coefficient of $R$ in terms of those of $P$ and $Q$ (without speaking of roots) boils down to expanding the internal coproduct $\Delta_{\times} e_{\ell}$ of the elementary symmetric function $e_{\ell} \in \Lambda$ (where $\Lambda$ is the ring of symmetric functions in infinitely many variables) in the basis $(e_{\lambda} \otimes e_{\mu})$ (with $\lambda$ and $\mu$ ranging over partitions) of the tensor product $\Lambda \otimes \Lambda$. This can be done: 
$$\Delta_{\times} e_{\ell} = \sum_{\lambda \vdash \ell} s_{\lambda} \otimes s_{\lambda^t},$$
where $\lambda^t$ denotes the transpose of an integer partition $\lambda$, and $s_{\lambda}$ is the Schur function corresponding to ${\lambda}$. 
The Schur functions can be written in terms of the elementary symmetrics using the von Nägelsbach-Kostka identities. This is probably going to be a mouthful of new words to you if you don't have an algebraic combinatorics background; I don't think anything simpler works, though. I have sketched a proof of the formula for $\Delta_{\times} e_{\ell}$ in the solution of Exercise 2.74(b) in Vic Reiner's notes on Hopf algebras in combinatorics, but this is probably not a good source on internal comultiplication.) This was about the characteristic polynomial; I can't say anything about the minimal polynomial.
On your question about $\psi$ being the identity matrix... well, that makes things simpler. Tensoring $\varphi$ with an $m \times m$ identity matrix is equivalent (in the sense that the results will be conjugate to each other) as taking the "block-diagonal" direct sum $\varphi^{\oplus m} \colon V^{\oplus m} \to V^{\oplus m}$. So the characteristic polynomial will be the $m$-th power of that of $\varphi$, and the minimal polynomial will be that of $\varphi$ (as long as $m \not= 0$).
Actually, my taking of internal coproducts was overkill. It is enough to look at the dual Cauchy identity:
$$
\prod_{i=1}^n\prod_{j=1}^m (1+x_iy_j)=\sum_{\lambda} s_{\lambda}(x_1,x_2,...,x_n)s_{\lambda^t} (y_1,y_2,...,y_m),$$
where the sum is over all partitions $\lambda$. (If you want, you can restrict it to partitions $\lambda$ whose largest part is $\leq m$ and whose length is $\leq n$; all other partitions contribute vanishing addends.) If you substitute $y_j t$ for $y_j$, you obtain
$$
\prod_{i=1}^n\prod_{j=1}^m (1+x_iy_jt)=\sum_{\lambda} s_{\lambda}(x_1,x_2,...,x_n)s_{\lambda^t} (y_1,y_2,...,y_m)t^{|\lambda|},$$
and you should recognize the left hand side as the "reversed" characteristic polynomial of the tensor product of a matrix whose "reversed" characteristic polynomial is $\prod_{i=1}^n (1+x_i t)$ and a matrix whose "reversed" characteristic polynomial is $\prod_{j=1}^m (1+ y_j t)$.
By "reversed" characteristic polynomial, I mean $\det(I_n+At)$ (where $A$ is the matrix and $n$ is its size), as opposed to the usual characteristic polynomial $\det(tI_n−A)$. The $x_i$'s and $y_j$'s are the coefficients of the respective reversed characteristic polynomials, or (up to sign and order) those of the usual characteristic polynomials, of $\varphi$ and $\psi$.
A: Assume $\dim(V)=m$, $\dim(W)=n$, $\lambda_1,\ldots,\lambda_m$ to be the eigenvalues of $\varphi$ and $\mu_1,\ldots,\mu_n$ those of $\psi$. As noticed in the comments, the idea is that the eigenvalues of $\varphi\otimes\psi$ are the products $\lambda_i\mu_j$ (see, e.g., this link). 
My answer covers a simple case: if the eigenvalues of $\varphi$ (or, equivalently, those of $\psi$) are known and different from zero, then the characteristic polynomial of $\varphi\otimes\psi$ is
$$
p_{\varphi \otimes \psi} 
= \prod_{i=1}^m \prod_{j=1}^n (x-\lambda_i \mu_j)
= \prod_{i=1}^m \lambda_i \prod_{j=1}^n \left(\frac{x}{\lambda_i}- \mu_j\right)
= \prod_{i=1}^m \lambda_i p_{\psi}\left(\frac{x}{\lambda_i}\right),
$$
where $p_{\psi}$ is the characteristic polynomial of $\psi$.
