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This is the explanation why $V$ is an $F[x]$-module:

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The book said "The linear map $T$ will enable us to make $V$ into an $F[x]$-module " why this happened?

Also, I did not understand the statement " $\circ$ denotes composition of functions (which make sense because the domain and codomain of $T$ are the same )", what is the relation between composition of functions, and domain & codomain? could anyone tell me please?

The book said that "The definition of the $F[x]$ action on $V$ is consistent with the given action of the field $F$ on the vector space $V$", but I can not see how, could anyone explain this for me please?

Could anyone explain the page for me in a clearer and simpler way please?

thanks!!

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    $\begingroup$ I think you should be more specific than "I don't understand this entire page. Someone explain it to me." Surely you get many parts of it, and can see the parts you don't get. If you don't, I'm skeptical about how much you can be helped... $\endgroup$ – rschwieb Oct 5 '17 at 16:20
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    $\begingroup$ Ok I do not get the idea of why $F[x]$ is a module ?@rschwieb $\endgroup$ – Intuition Oct 5 '17 at 18:15
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    $\begingroup$ @rschwieb Also, I did not understand the statement " $\circ$ denotes composition of functions (which make sense because the domain and codomain of $T$ are the same )", what is the relation between composition of functions, and domain & codomain? could anyone tell me please? $\endgroup$ – Intuition Oct 6 '17 at 5:08
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    $\begingroup$ Also @rschwieb The book said "The linear map $T$ will enable us to make $V$ into an $F[x]$-module " why this happened? $\endgroup$ – Intuition Oct 6 '17 at 6:58
  • $\begingroup$ Expanded my answer based on those questions. Might be too late but I wanted to give my version for completeness of my answer. $\endgroup$ – rschwieb Oct 6 '17 at 10:49
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I'll illustrate with an example, and perhaps it helps.

Take $F = \Bbb R$, so $R$ is the ring of polynomials with real coefficients, and let $V = \Bbb R^2$, so that our vector space is the plane with regular vector addition and scalar multiplication. Finally, let $T$ be rotation anti-clockwise by $90^\circ$, represented in the standard basis, as you probably know, by the matrix $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ which I shall also call $T$, by abuse of notation.

$V$ is a module over $\Bbb R$ via the regular scalar multiplication. Now we make $V$ into an $\Bbb R[x]$ module by defining the following module multiplication: $$ p\cdot v = p(T)v $$ where the right side is regular matrix multiplication. An example of this multiplicaion might make things even clearer: If $p(x) = x^2 + x + 1$, and $v = (1, 1)^t$, then we have $$ p\cdot v = (T^2 + T + 1)v = T^2v + Tv + v\\ = \begin{pmatrix}-1\\-1\end{pmatrix} + \begin{pmatrix}-1\\1\end{pmatrix} + \begin{pmatrix}1\\1\end{pmatrix}\\ = \begin{pmatrix}-1\\1\end{pmatrix} $$

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  • $\begingroup$ The book said "The linear map $T$ will enable us to make $V$ into an $F[x]$-module " why and how this happened? $\endgroup$ – Intuition Oct 6 '17 at 4:50
  • $\begingroup$ Also, I did not understand the statement " $\circ$ denotes composition of functions (which make sense because the domain and codomain of $T$ are the same )", what is the relation between composition of functions, and domain & codomain? could anyone tell me please? $\endgroup$ – Intuition Oct 6 '17 at 5:08
  • $\begingroup$ Sorry, you already explained how :) $\endgroup$ – Intuition Oct 6 '17 at 5:39
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    $\begingroup$ @Intuition If you want to compose two functions $f$ and $g$ to make a new function $f\circ g$, then you have to make sure that whatever you get out of $g$ (i.e. the codomain of $g$) is something you can apply $f$ to (i.e. is in the domain of $f$). In this case, what you get out of $T$ (which is a vector in $V$) is something that it makes sense to apply $T$ to. That means you can safely compose $T$ with $T$. $\endgroup$ – Arthur Oct 6 '17 at 6:36
  • $\begingroup$ Is there is a relation between the above prove and the proof that $R[x_{1} , x_{2}, \dots , x_{n}]$ is an $R$- module? $\endgroup$ – Intuition Oct 6 '17 at 6:52
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Ok I do not get the idea of why $F[x]$ is a module ?

The contents of the page you included explain why $V$ is an $F[x]$-module. It does so by defining an action $F[x]\times V\to V$.

$F[x]$ is a module over $F[x]$, but that does not seem like the main point of this page.

what is the relation between composition of functions, and domain & codomain?

If $f:A\to B$ and $g:B\to C$, then you can compose them to become $g\circ f:A\to C$ because the codomain of $f$ matches the domain of $g$. If you don't have this condition, the composition doesn't make sense. If you want to compose a map with itself, well then you'd better have $T:V\to V$ like we have here so that things line up.

"The linear map $T$ will enable us to make $V$ into an $F[x]$-module " why this happened?

$V$ was already a vector space, so we knew how $F$ was acting on $V$. To make $x$ act on $V$, we're selecting a linear transformation $T$ to tell us what $x$ does to $V$.

After you decide what $x$ does to $V$, then you have no choice about what $x^n$ does to $V$, because it must be the same thing as repeatedly applying $T$ $n$ times. By linearity, this completely determines what $\sum_{i=1}^n \alpha_ix^i$ does to $V$.

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