Writing an explicit map and making a vector space defined over field $F$ into a $F[x ,y]$-module

Example 10.3.13 ($F[x]$-modules). Let $F$ be a field and let $R$ be the polynomial ring $R = F[x]$. We propose to determine what left $F[x]$-modules may be. Let $M$ be a left $F[x]$-module.

First note that $F$ (as a subring of $F[x]$) must act on $M$ according to the left $R$-module axioms. Thus $M$ is a vector space over the field $F$. Now consider the action of the element $x$ on $M$. The axioms for a left-module show that the element $x$ behaves as a linear transformation $T$ on the $F$-vector space $M$. Furthermore, the axioms impose no further restrictions on the linear transformation $T$ we use for $x$. Now for any polynomial $p(x) = a_d x^d + \cdots + a_1 x + a_0 \in F[x]$ and for any $v \in M$,

$$ \begin{aligned} P(x) \cdot v &= (a_d x^d + \cdots + a_1 x + a_0) \cdot v \\ &= a_d \overbrace{T(T \cdots T}^{\text{$d$ times}}(v) \cdots )) + \cdots + a_1 T(v) + a_0 v. \end{aligned} $$

Consequently, an $F[x]$-module consists of two data: an $F$-vector space $V$ and a linear transformation $T : V \to V$. The above equation gives teh action of any polynomial on a vector in $V$.

I have a questions from an exercise which is from the book Abstract Algebra Structures and Applications pg 495. The question reads as follows:

Let $F$ be a field and consider the polynomial ring $F[x,y]$.

(a) Prove that an $F[x,y]$-module consists of vector space $V$ over $F$ along with commuting linear transformation $T_1$ and $T_2$.

(b) Explicitly describe the action of a polynomial $p(x,y)$ on a vector in $V$

Can I get some clarification on what the first question is asking as well, what the difference between the first and second question are? Is it asking me to show that $V$ as a vector space acted on by the polynomial ring $F[x,y]$ can be make into a module?

As for making the vector space $V$ over $F[x,y]$ into a module, don't I need to worry about the dimension of $V$ since I will need evaluation maps to do so. I included a screenshot of how the author of the text does it in an example for the case of making a vector space $V$ into a $F[x]$ module. I think for the exercise question, the author expects me to give an explicit mapping. So if I disregard the issue of dimension and just attempt a solution like the way the author has done in the included image, I would apply $F[X]$ acting on $V$ and then apply $[y]$ acting on the action of the action of $F[x]$ on $V$. In mathematical notation, it would be as follows:

Given a vector space $V$, polynomial rings $F[x]$ and $F[x,y]$. Let $v\in V$, $\alpha_{i}\in F[x]$ with $\alpha_{i}=\sum_{r_{i}=0}^{i}a_{{r_{i}}{i}}x^{r_{i}}$, define the action of $F[x]$ on the vector space $V$ over $F$ by:$\alpha_{i}\cdot_{T_{1}}v=\alpha_{i}(T_{1})v=\sum_{r_{i}=0}^{i}(a_{{r_{i}}{i}}{T_{1}}^{r_{i}})v=\sum_{r_{i}=0}^{i}a_{{r_{i}}{i}}{T_{1}}^{r_{i}}(v).$

Again let $v\in V$, $p\in F[x,y]$ with $p=\sum_{i=0}^{n}\alpha_{i}y^{i}$ and $\alpha_{i}\in F[x]$, then the action of $F[x,y]$ on $V$ can be defined by:
$p\cdot_{T_{2}}v=p(T_{2})(v)=\sum_{{i}=0}^{n} (\alpha_{i}{T_{2}}^{i})v= \sum_{{i}=0}^{n} (\alpha_{i}{T_{2}}^{i}(v))=\sum_{{i}=0}^{n}(\sum_{r_{i}=0}^{i}a_{{r_{i}}{i}}{T_{1}}^{r_{i}}(v)){T_{2}}^{i}(v)=\sum_{{i}=0}^{n}\sum_{r_{i}=0}^{i}a_{{r_{i}}{i}}{T_{1}}^{r_{i}}(v){T_{2}}^{i}(v)=\sum_{{i}=0}^{n}\sum_{r_{i}=0}^{i}a_{{r_{i}}{i}}({T_{1}}^{r_{i}}{T_{2}}^{i})(v)$
with ${T_{1}}^{r_{i}}{T_{2}}^{i}={T_{2}}^{r_{i}}{T_{1}}^{i}$ and ${T_{k}}^{p}$ to mean $T_{k}$ composed with itself $p-$times, for $k=1,2$

Can someone look over my work to see if the explicit map I constructed or any of the notations I used for describing how the polynomials rings action on vector space $V$ are correct and which parts of (a) and (b) have I answered, and the parts I have not given an adequate answer, what is it I still need to or have not shown in order to answer the question. Thank you in advance.

It looks like you have the right flavor of idea, though honestly your notation is so condense that it's hard to read exactly what you're saying. I'll give some hints as to what you should do, and hopefully you can tell me if this is what you've already been doing :P

First, you ask for the difference between parts (a) and (b). They're definitely related, and I think the best thing to do is to try and understand what $F[x,y]$-modules are. Then you'll get an answer to parts (a) and (b) at roughly the same time. After all, they're asking for two different aspects of the characterization of $F[x,y]$-modules.

Recall from your example $10.3.13$ that an $F[x]$-module is really a vector space $V$ plus a linear map $T : V \to V$. Part (a) is asking you to show that an $F[x,y]$-module is really a vector space $V$ plus two linear maps $T_1$ and $T_2 : V \to V$. But they can't be any old linear maps! They must commute. That is, $T_1 T_2 v = T_2 T_1 v$ for every $v \in V$. It's worth thinking about why this is.

Also recall from the same example that we get a computation rule from the above characterization. That is, if you give me a vector space $V$ and a linear map $T : V \to V$, then I know what $p(x) \cdot v$ is! Indeed, if we take a concrete example, say, $x^2 - 3x + 1$. Then

$$(x^2 - 3x + 1) \cdot v = T^2 v - 3Tv + v$$

what we do for other polynomials is hopefully obvious from this example and the discussion in your example.

Part (b) is asking for a similar computation rule for $F[x,y]$-modules. That is, if $T_1$ and $T_2 : V \to V$ and you give me a polynomial $p(x,y)$, say, $x^2 + xy + 3y + 2$, what is $(x^2 + xy + 3y + 2) \cdot v$? Now, in the example you had a polynomial in one variable and one linear map $T$. Now we have a polynomial in two variables and two linear maps $T_1$ and $T_2$.... Can you guess what the computation rule should be? I'll include this particular example under the fold if you need a hint:

$$(x^2 + xy + 3y + 2) \cdot v = T_1^2 v + T_1 T_2 v + 3 T_2 v + 2v$$

Now with this in mind it might be clearer we we need to know that $T_1 T_2 = T_2 T_1$. After all, we know that in $F[x,y]$ we have $xy = yx$... Can you use this equation, plus the axioms for modules, to show that if $T_1$ and $T_2$ are linear maps making $(V, T_1, T_2)$ a $F[x,y]$-module, then we must have $T_1 T_2 = T_2 T_1$?

I hope this helps ^_^