I came across questions in the free module section of my abstract algebra text. In the text, the notation $End_{R}(V)$ denotes the set of all $R$-module endomorphisms of $M$. Algebra: Abstract and Concrete, exercise 8.1.9 on p358 in the attached picture of the linked text Onto the question:
Let $V$ be a finite dimensional vector space over a field $K$. Let $T\in End_{K}(V)$. Give $V$ the corresponding $K[x]$-module structure defined by $\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$ Show that $V$ is not free as a $K[x]$-module.
In this question, I don't understand how in $\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$ affects whether $V$ can be a free $K[x]$-module. If I take a finite basis $B$ for $V$, where $B=\{v_1, v_2,...v_n\}$, with coefficients $\alpha_i \in K$, then would it be that each of the $\alpha_i x^i v_{i}$ term in the identity $\alpha_i x^i v_{i} = \alpha_{i}T^{i}(v_i)$, the coefficients $\alpha_{i}$ or $\alpha_{i}x^{i}$ might not equal to zero? Actually, I am assuming the $v$ in the definition $\alpha_i x^i v = \alpha_{i}T^{i}(v)$ refers to basis elements from $B$, but I am not sure where the $x^{i}$ is suppose to come from. Is it from the vector space $V$ or from the field $K$ along with the $\alpha_i$.
Thank you in advance.
In the statement $x^i$ means neither an element of $V$ nor $K$. Let me explain the situation: $V$ is a $K$-vector space, so in particular is an abelian group. This means that we can turn $V$ into a module over the ring $R := K[x]$ by specifying what $p \cdot v$ is for each $p \in K[x]$ and $v \in V$ (and then checking that the right identities are satisfied). Recall that $K[x]$ is just the set of all polynomials in the formal variable $x$ and with coefficients in $K$. So then this is what your book does: for any polynomial $p = \sum_i \alpha_i x^i \in K[x]$ and any $v \in V$, we declare that $$ p \cdot v := \sum_i \alpha_i T^i(v). $$ Since $T^i(v)$ is again a vector in $V$ for each $i$ and this is a finite sum, the result $p \cdot v$ is again an element of $V$, and so we have a well-defined module action.
Now that we know what the module structure is (and in particular that $x$ is a formal variable), we can think about the question: we are supposed to show that $V$ is not a free $K[x]$-module. Well, suppose $V$ was free (as a $K[x]$-module) with basis $\mathcal{B} := \{v_1, \ldots, v_n\} \subset V$. From linear algebra (the Cayley-Hamilton theorem) we know that every $T \in \operatorname{End}_K(V)$ is a root of its own (nonzero) characteristic polynomial $p \in K[x]$. Thus $p \cdot v_1 = p(T)(v_1) = 0$, which violates the fact that $\mathcal{B}$ is supposedly $K[x]$-linearly indepedent. Hence the $K[x]$-basis $\mathcal{B}$ cannot exist.
Finally let's address the question of why the action of $K[x]$ on $V$ matters for freeness. The point is that to check freeness of $V$ as a $K[x]$-module we need to check that any basis $\mathcal{B} \subset V$ is both spanning and linearly indepednent over $K[x]$! As you can see above for any $T \in \operatorname{End}_K(V)$ we can always find a $p \in K[x]$ such that $p \cdot v = 0$ for all $v \in V$, which implies that a basis for $V$ over $K[x]$ can never exist. On the other hand if we wanted to consider $V$ as an ordinary $K$-module of course $V$ does have a basis (indeed, many) since this is just what it means to be an ordinary basis for a vector space over $K$.
Note that any ordinary $K$-basis $\mathcal{B}$ for $V$ is $K$-spanning so of course is $K[x]$-spanning too, but the above implies that such a basis couldn't be $K[x]$-linearly independent.