Not really sure what a $k[x]$-module is.

Question

Let $A = k[x]$ where $k$ is a field. Then an $A$-module is just a $k$-vector space $V$ equipped with a $k$-linear map $\widehat{x}: V \to V$.

This statement is not clear to me. Could anybody help elaborate it for me?

Answer 1

I am going to write $T$ instead of $\widehat{x}$ everywhere.

Given a $k$-vector space $V$ together with a linear transformation $T$, we may give $V$ the structure of a $k[x]$-module as follows: $k$ acts on $V$ by scalar multiplication, and $x$ acts on $V$ by $T$. This tells us how anything in $k[x]$ acts since everything in $k[x]$ is a sum of products of scalars and $x$'s.

Conversely, given a $k[x]$-module $M$, we see that $M$ comes with an action of $k$, so is a vector space. Write $T$ for the map $M \to M$ given by multiplication by $x$. Then, since $k[x]$ is commutative, $T$ is $k$-linear for the aforementioned $k$ vector space structure.

These two processes are inverse to each other. Therefore, $k[x]$-modules "are the same thing as" vector spaces together with a linear transformation $T$ (where "are the same thing as" means: there is a canonical identification of these two categories).

Answer 2

By the usual definition, an $A$-module is an abelian group $V$ together with an action of $A$ on it. The action is given by a ring homomorphism $A\to \operatorname{End}(V)$.

In the special case that $A=k[x]$, recall that the polynomialk ring is defined by a universal property: $k[x]$ is a ring together with a homomoprhism $i\colon k\to k[x$ and a special element $x\in k[x]$, such that for any ring homomorphims $f\colon k\to Y$ and choice of element $\hat x\in Y$, there exists exactly one ring homomorphism $h\colon k[x]$ such that $h\circ i=f$ and $h(x)=\hat x$.

Thus the action of $k[x]$ on $V$ is given by a ring homorphism $k\to \operatorname{End}(V)$ and a specific endomorphism $\hat x\in\operatorname{End}(V)$. Now as $k\to \operatorname{End}(V)$ is an action of a field on an abelian group $V$, we can just say that $V$ is a $k$-vector space. And there we are: an $A$-module is a $k$-vector space $V$ together with a $k$-linear map $\hat x\colon V\to V$.

Answer 3

Let $V^+$ denote the underlying Abelian group of $V$.

Viewing $k$ simply as a commutative ring with a $1$ (the field property of $k$ is merely a distraction here), a $k[x]$-module is ("essentially") a ring homomorphism $k[x] \to \operatorname{End}(V^+)$.

By the universal property of polynomial rings - again, the field property of $k$ plays no part - there is exactly one such homomorphism that (a) extends a given ring homomorphism $k \to \operatorname{End}(V^+)$ and (b) maps $x$ to an arbitrarily given element of $\operatorname{End}(V^+)$.

But a ring homomorphism $k \to \operatorname{End}(V^+)$ is (again, "essentially") a $k$-module, i.e., because $k$ happens to be a field, it is a $k$-vector space.

Thus, a $k[x]$-module is determined by two "data": (a) a $k$-vector space, which it extends, plus (b) an arbitrary element of $\operatorname{End}(V^+)$, representing multiplication by the scalar $x$.