Paul Garret's Proof of the Cayley-Hamilton Theorem I am trying to understand the proof of Cayley-Hamilton Theorem given in Paul Garrett's notes.
We have a finite dimensional vector space $V$ over a field $k$ and are given a linear operator $T\in \mathcal L(V)$.
The proof on pg 431 in the above link starts out as:
The module $V\otimes_k k[x]$ is free of rank $\dim_k(V)$ over $k[x]$. Also, $V$ is also a $k[T]$ module by the action $v\mapsto Tv$. (I understand this much).
Now the next line reads:
So $V\otimes_k k[x]$ is a $k[T]\otimes_k k[x]$ module.

This I do not understand.

What is the general fact at play here?
EDIT: To expand on my (complete) lack of understanding, I do not see how are we giving a $k[T]\otimes_k k[x]$ module structure to $V\otimes_k k[x]$.
And of course, I am looking for a 'general principle' at work here.
For example, when we said that $V\otimes_k k[x]$ is a $k[x]$-module of rank $\dim_k(V)$, what we are using is the following: We have a natural injection $i:k\to k[x]$. So we can extend the scalars on the $k$-module $V$ and get a $k[x]$-module $V\otimes_k k[x]$.
Since $V\otimes_k k[x]\cong k[x]^n$, where $n=\dim_k(V)$, we also know that the rank of $V\otimes_k k[x]$ as a $k[x]$-module is same as $\dim_k(V)$.
 A: $V$ is a $k[T]$-module.  
$V \otimes $ Something is a ($k[T] \otimes$ Something) - module.  
A: If you have an $N\times N$ matrix $A=[a_{n,m}]$ over a field, then the cofactor expansion of the determinant of $A$ gives you
$$
                \mbox{adj}(A)A=\mbox{det}(A)I,
$$
where $\mbox{adj}(A)$ is the adjunct or adjugate matrix consisting of the cofactors of $A$. Therefore,
$$
               \mbox{adj}(\lambda I-A)(\lambda I-A)=p(\lambda)I
$$
where $p(\lambda)=\mbox{det}(\lambda I - A)$ is the characteristic polynomial of $A$. You can write this as
$$
              (A_{0}+\lambda A_{1}+\cdots+\lambda^{n-1}A_{n-1})(\lambda I-A)=p(\lambda)I,
$$
where $A_n$ are $N\times N$ coefficient matrices. The polynomial
$$
              Q(\lambda) = A_0+\lambda A_1 + \cdots +\lambda^{n-1}A_{n-1}
$$
has coefficient matrices that may or may not commute with $A$. However, the coefficient matrices of $\lambda I -A$ do commute with $A$. Whenever you have polynomials $Q(\lambda)R(\lambda)=S(\lambda)$ and the coefficients of $R$ commute with $A$, then $(QR)|_{A}=Q|_{A}R|_{A}=S|_{A}$, where the evaluation is done on the right, as opposed to being evaluated on the left. In this case,
$$
              Q|_{A}(\lambda I-A)|_{A}=p(A) \\
                    Q|_{A} 0 = p(A).
$$
That's the basic idea behind it.
