Determinant of Kronecker product, approach via $F[x]$-modules Remark It's easy to see that every linear operator $f\colon M\to M$, where $M$ is a $R$-module, might be seen as a $R[x]$-module $M_f$ via homomorphism $R[x]\to \operatorname{End}_{R} (M)$ s.t. $r\mapsto r \ \forall r\in R, x\mapsto f$ (more explicit, the «extension of scalars» occurs, elements from $R$ act the same manner as in $M$, $xm:=f(m)$). In the case if $R$ is a field and $M$ is a finite dimensional vector space it's possible to construct linear operator theory very comfortable using classification theorem of finitely generated modules over PID.
Now let $F$ be a field, $V, W$ be a finite dimensional vector spaces over $F,\ f\in \operatorname{End}_F (V), \ g\in \operatorname{End}_F (W)$. It shall be great if we could see $f\otimes g$ (tensor product of two linear maps) as a module $V_f\otimes_{F} W_g$.
If this idea is correct, then we can easily find $\det(f\otimes g)$, for instance. Let us explore the most simple case, assume $$[1]\otimes [1]\in \frac{F[x]}{\langle x-\lambda \rangle }\otimes_F \frac{F[x]}{\langle x-\lambda' \rangle }.$$ Multiplication by $x\otimes x$ gives us $$[x]\otimes[x]=[\lambda]\otimes[\lambda']=\lambda\lambda'([1]\otimes [1]),$$ therefore $\lambda\lambda'$ is an eigenvalue of $f\otimes g$.
And then it's easy to improve the idea above for evaluating the determinant keeping in mind that the algebraic closure (or, at least, the splitting field of characteristic polynomial) always exists.
The question is. Is that idea correct?
Thanks for paying attention.
UPD. I've fixed some typos and the mistake mentioned in the answer.
 A: You need to tensor over $F$, not over $F[x]$ to get the Kronecker product. The action of $x$ on $V_f\otimes_F V_g$ is defined on elementary tensors as $x(v \otimes w):=xv \otimes xw$ and extended linearly. Other than that, your idea is correct.
A: Three comments.

*

*You don't need to know anything about modules to compute that if $v, w$ are eigenvectors of two linear maps $f, g$ with eigenvalues $\lambda, \mu$ then $(f \otimes g)(v \otimes w) = f(v) \otimes g(w) = \lambda v \otimes \mu w = (\lambda \mu) v \otimes w$, so $v \otimes w$ is an eigenvector of $f \otimes g$ with eigenvalue $\lambda \mu$.


*This does not immediately imply that $\det(f \otimes g) = \det(f) \det(g)$ because $f, g$ may not be diagonalizable (even after passing to the algebraic closure) and this argument doesn't imply that multiplicities match up. However there are standard arguments that let you deduce the general case from the diagonalizable case, e.g. Zariski density.


*We know that if $R$ is a commutative ring then the tensor product of two $R$-modules over $R$ is another $R$-module. This is not the tensor product relevant to this question, which happens over the base field and not over $F[x]$, and you might wonder what kind of tensor product this is instead. The answer is that this tensor product comes from a bialgebra structure on $F[x]$, namely the comultiplication $x \mapsto x \otimes x$. Different bialgebra structures lead to different $F[x]$-module structures on the tensor product of two $F[x]$-modules over $F$; for example, the bialgebra structure $x \mapsto x \otimes 1 + 1 \otimes x$ leads to the Kronecker sum $f \otimes 1 + 1 \otimes g$, which appears for example if we think of $F[x]$ as the universal enveloping algebra of the $1$-dimensional Lie algebra.
