Question about finite dimensional vector space over a field $K$ and free $K[x]$-module

Solution 1:

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 $K[x]$. 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.

Solution 2:

And $R$-module $M$ is free if it has a subset $S\subset M$ that generates $M$ over $R$, and that is linearly independent over $R$. The $R$-module structure of $M$ matters here: For this particular $K[x]$-module structure on $V$, even any singleton is not linearly independent, because for $f\in K[x]$ and $v\in V$ $$f\cdot v=0\qquad\not\Longrightarrow\qquad f=0.$$ That is to say, for every $v\in V$ there exists some nonzero $f\in K[x]$, say $f=\sum_i\alpha_ix^i$, such that $$0=f\cdot v=(f(T))(v)=\sum_i\alpha_iT^i(v).$$ In fact, a much stronger claim can be made; there exists some nonzero $f\in K[x]$ such that $f\cdot v=0$ for all $v\in V$. This is a polynomial in $K[X]$ that satisfies $(f(T))(v)=0$ for all $v\in V$, or equivalently, that satisfies $f(T)=0$. This is of course the minimal polynomial of $T$, or any multiple of it.

Anyway, we see that the only subset of $V$ that is $K[x]$-linearly independent (w.r.t. this particular $K[x]$-module structure) is the empty set. In particular $V$ is free over $K[x]$ if and only if $V=0$.