I am trying to understand the proof of Cayley-Hamilton Theorem given in Paul Garrett's [notes](http://www.math.umn.edu/~garrett/m/algebra/notes/28.pdf).

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.*

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> This I do not understand.
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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 $\dim\_k(V)$.