A finite dimensional vector $V$ is not free as an $\operatorname{End}_{K}(V)$-module I came across questions in the free module section of my abstract algebra text. As on p358 of the text Algebra: Abstract and Concrete, the notation $\operatorname{End}_{R}(M)$ denotes the set of all $R$-module endomorphisms of $M$. On to the question:

Let $V$ be an $n$-dimensional vector space over a field $K$, with $n>1.$  Show that $V$ is not free as an $\operatorname{End}_{K}(V)$ module.

I don't understand the difference between a "free"-module vs a '"free" $End_{K}(V)$ module'.  I know that any vector space $W$ over a field $F$ has a basis and hence it is a free $F$-module.  In the case of the stated $n$-dimensional vector space $V$ in the question, how does defining an endomorphism map $\phi:V\rightarrow V$ given by $\phi(rm)=r\phi(m)$ for all $r\in K$ affects whether $V$ can still being a free module.  Or is it because once I defined some sort of endomorphism map $\phi$ between $V$ to itself, whether $V$ can still enjoy being a free module, or a free $\operatorname{End}_{K}(V)$ module?  I think there is something subtle that I am either not understanding or I am missing to see.  I also understand that a vector space over a field is always a free module because it has a basis and it is defined with scalars from a field which is also an integral domain and the coefficients of the elements of the basis are zero divisors.
Thank you in advance.
 A: As you rightly say any vector space $V$ is a free module over $K$. The problem refers to now considering $V$ as a module over the different (now noncommutative) ring $\operatorname{End}_K(V)$. A basis for $V$ as a $K$-module need not be a basis for $V$ as an $\operatorname{End}_K(V)$-module, and that's why $V$ may be free over $K$ and not over $\operatorname{End}_K(V)$.
Now on to the actual problem---this follows from a general fact about $R$-modules. Namely, we have:
Lemma. Let $R$ be a ring and let $M$ be an $R$-module. If for all $m \in M$ there is a nonzero $r \in R$ such that $r \cdot m = 0$, then $M$ is not a free $R$-module.
Proof. For a contradiction suppose that $\mathcal{B} = \{m_1, \ldots, m_n\}$ is an $R$-basis for $M$. Then by hypothesis there exists a nonzero $r \in R$ such that $r \cdot m_1 = 0$: this contradicts $R$-linear independence of $\mathcal{B}$, as desired.
Now given this lemma, your question is answered quickly: we just need to find for each $v \in V$ a nonzero $\phi \in \operatorname{End}_K(V)$ such that $\phi(v) = 0$. (Note that if $\dim V = 1$ then this is impossible, so that's why it is a hypothesis for your problem.) First note that if $v = 0$ then we can choose $\phi$ to be any nonzero endomorphism of $V$. Otherwise $v \not = 0$, and we can extend the set $\{v\}$ to a basis $\mathcal{B} = \{v, b_1, \ldots, b_n\}$ for $V$. Now define a linear map $\phi$ by declaring that $\phi(v) = 0$ and $\phi(b_i) = 1$ for all $1 \leq i \leq n$. Since $n = \dim V > 1$ then $\phi$ is nonzero because $\phi(b_1) = 1$, but on the other hand $\phi(v) = 0$. We conclude by the lemma that $V$ is not a free $\operatorname{End}_K(V)$-module, as you desire.
