Questions on what a $k[x]$-module is, followup.

Question

This is a followup to my question here. I have tried reading some of the answers, and I don't really understand what they are saying in places.

1. 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 $A \to \text{End}(V)$.

   I don't really understand why the action is given by a ring homomorphism $A \to \text{End}(V)$. Could anybody elaborate on this?

2. 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.

   How does $x$ act on $V$ by $T$? And why does this tell us that how anything in $k[x]$ acts since everything in $k[x]$ is a sum of products of scalars and $x$'s?

3. 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.

   Why can we write $T$ for the map $M \to M$ given by multiplication by $x$? And why does $k[x]$ being commutative imply that $T$ is $k$-linear?

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

   One point of confusion with this is as follows. So is a specific $A$-module equipped with a fixed linear map $\widehat{x}$ that corresponds to a specific element in $k[x]$? Or is $\widehat{x}$ allowed to vary as to correspond to all elements of $k[x]$? Or am I thinking of this in the wrong way and does this $\hat{x}$ correspond to $k[x]$ aggregate and not an element or elements in it?

Thanks in advance!

Answer 1

1. This is because, if $a\in A$, for any $u, v\in V$, we have $a(u+v)=au+av$ and $\cdot 0=0$, by two of the axioms which make up the definition of an $A$-module. So scalar multiplication $m_a$ on $V$ is a group endomorphism, and we have a map $$\DeclareMathOperator{\End}{End}m \colon A \longrightarrow \End(V),\quad a \longmapsto(m_a\colon u\mapsto au)$$ This map is a ring homomorphism, because $m_1=\operatorname{id}_V$, the unit element of $\End(V)$ and $m_{a+b}=m_a+m_b$, which are translations of the axioms $\;1\cdot u=u$ and $\;(a+b)u=au+bu$.
2. Don't forget $x$ is not a variable, but an indeterminate: it does not represent an element of $k$, and exists by itself. So it's better to use the formal notation X. This means we define $\; X\cdot u$ as $T(u)$. As we have a ring homomorphism, there follows $\;X^2\cdot u=(T\circ T)(u)=T\bigl(T(u)\bigr)$, and similarly $$X^k\cdot u=(\underbrace{T\circ T\circ \dotsm\circ T}_{k\;\text{factors}})(u) $$ For a general polynomial, the formula follows by linearity.
3. If multiplication by $X$ is $k$-linear, so it is a $k$-endomoprphism. Why couldn't denote multiplication by $X$ as $T$ when we view it as a vector space endomorphism? Now this pultplication is $k$-linear because, if $\lambda\in k$ and $m\in M$, by the axioms of modules, and commutativity of $k[X]$, $$T(\lambda m)\stackrel{\text{def}}{=}X(\lambda m)=(X\lambda) m=(\lambda X) m =\lambda (Xm)=\lambda T(m).$$
4. The $k$-linear map $\widehat X$ corresponds to multiplication by $X$. This because a $k$-algebra homomorphism from the ring of polynomials $k[X]$ into a $k$-algebra $E$ is uniquely determined by the image of $X$.

Answer 2

This answer addresses question 1 only.

Temporarily forget about the operations of addition on the sets $A$ and $V$.

That is, think of $V$ for the time being as just a set, and $A$ as just a multiplicative monoid.

An action of $A$ on $V$ can be described in two ways, and they are equivalent, in s sense to be explained.

There is an operation of "scalar multiplication", $A \times V \to V$, $(\alpha, v) \mapsto \alpha{v}$, satisfying two postulates:

• $1v = v$, for all $v \in V$.

• $(\alpha\beta)v = \alpha(\beta{v})$, for all $v \in V$ and all $\alpha, \beta \in A$.

We can curry this function, obtaining a function $\mu: A \to (V \to V)$, where $(V \to V)$ denotes the set of all functions $V \to V$ - if that's not too confusing! - defined by:

• $(\mu(\alpha))(v) = \alpha{v}\ (\alpha \in A$, $v \in V$).

If $\iota$ denotes the identity function $V \to V$, we then have

• $\mu(1) = \iota$,

• $\mu(\alpha\beta) = \mu(\alpha) \circ \mu(\beta)$, for all $\alpha, \beta \in A$.

That is, $\mu$ is a homomorphism of the monoid $A$ into the multiplicative monoid $(V \to V)$ of functions $V \to V$ - sorry about the notation, again! - under the associative operation of function composition, $\circ$.

Conversely, let $\mu: A \to (V \to V)$ be any monoid homomorphism, and define scalar multiplication, $A \times V \to V$, $(\alpha, v) \mapsto \alpha{v}$, by:

• $\alpha{v} = (\mu(\alpha))(v)\ (\alpha \in A$, $v \in V$).

Then the two postulates for scalar multiplication given above are satisfied.

Moreover, if we pass from "scalar multiplication" to "monoid homomorphism" and back again, using these two processes, we get back to the same operation that we started with; ditto, if we pass from "monoid homomorphism" to "scalar multiplication" and back again.

So, there is a kind of equivalence between these two ways of describing an action of a multiplicative monoid $A$ on a set $V$.

If we now recall the operations of addition on $A$ and $V$, we can define the ring $\operatorname{End}(V^+)$ of functions $V \to V$ which respect the operation of addition on $V$; and at the same time, we can specialise to operations of scalar multiplication, $A \times V \to V$, $(\alpha, v) \mapsto \alpha{v}$ satisfying two further postulates:

• $\alpha(v + w) = \alpha{v} + \alpha{w}$, for all $v, w \in V$ and $\alpha \in A$,

• $(\alpha + \beta)v = \alpha{v} + \beta{v}$, for all $\alpha, \beta \in A$ and $v \in V$.

That is:

• $\mu(\alpha) \in \operatorname{End}(V^+)$, for all $\alpha \in A$,

• $\mu(\alpha + \beta) = \mu(\alpha) + \mu(\beta)$, for all $\alpha, \beta \in A$.

Therefore, under the process that has already considered for monoid actions in general, the monoid homomorphism $\mu: A \to (V \to V)$ range-restricts to a monoid homomorphism $A \to \operatorname{End}(V^+)$, which is in fact a ring homomorphism. Conversely, given a ring homomorphism $\mu: A \to \operatorname{End}(V^+)$, and defining scalar multiplication in the same general way as before, we find that the two new postulates are satisfied.

So, the general equivalence between two ways of specifying an action of $A$ on $V$ still holds, and when specialised in the way just described, it becomes an equivalence between: (1) operations of scalar multiplication satisfying all the four postulates listed, and (2) ring homomorphisms $A \to \operatorname{End}(V^+)$.