TITLE: Finding a presentation matrix for an $F[t]$-module
QUESTION [2 upvotes]: Here is an exercise from Artin:

Let $V$ be an $F[t]$-module, and let $\mathbf B=(v_1,\dots,v_n)$ be a basis for $V$ as $F$-vector space. Let $B$ be the matrix of $T$ with respect to this basis. Prove that $A=tI-B$ is a presentation matrix for the module.

(Note that $T$denotes the linear operator $V\to V$ defined by $T(v)=tv$.)
Artin defines a presentation matrix as follows:

Left multiplication by an $m \times n$ matrix defines a homomorphism
  of $R$-modules $A: R^n \rightarrow R^m$. Its image consists of all
  linear combinations of the columns of $A$ with coefficients in the
  ring, and we may denote the image by $AR^n$. We say that the quotient
  module $V=R^m/AR^n$ is presented by the matrix $A$. More generally,
  we call any isomorphism $\sigma: R^m/AR^n \rightarrow V$ a
  presentation of a module $V$, and we say that the matrix $A$ is a presentation matrix for $V$ if there is such an isomorphism.

Back to the problem, I'm not even sure what I'm supposed to show. Presentation matrices are defined for quotient modules. Why is an arbitrary $F[t]$-module $V$ a quotient module? If $V$ were finitely generated, this would certainly be true since there would be a surjective $F[t]$-module homomorphism $(F[t])^n\to V$, and we could apply the First Isomorphism Theorem. So, why is a presentation matrix defined for $V$ in question, what exactly do I need to show, and at least how to get started?

Edit: alright, indeed, as @Mohan pointed out in the comments, since $V$ is generated by $\mathbf B$ as an $F$-vector space, then it is also generated by $\mathbf B$ as an $F[t]$-module. So there is a surjective module homomorphism $\phi: F[t]^n\to V$ given by $X\mapsto \mathbf B X$, where the RHS is the product of the row $\mathbf B$ by the column $X$. So $V$ is isomorphic to $F[t]^n/\ker \phi$. If we show that $\ker \phi= AF[t]^n$ where $AF[t]^n$ stands for the image of the map $A: F[t]^n\to F[t]^n$ given by $X\mapsto AX$ (here I denote the map and the matrix by the same letter $A$), then we will be done.

REPLY [1 votes]: Set $R:=F[t]$ and identify $V$ to $F^n$ thanks to its $F$-basis. Consider the $R$-linear maps 
$$
R^n\xrightarrow{tI-B}R^n\xrightarrow\phi F^n\to0,\quad(\ast)
$$ 
where $\phi$ is defined by 
$$
\phi\left(\sum t^ix_i\right)=\sum B^ix_i
$$ 
for $x_i\in F^n$. It suffices to prove that $(\ast)$ is an exact sequence. The surjectivity of $\phi$ and the equality $\phi\circ(tI-B)=0$ are clear. 
Let 
$$
\sum_{i=0}^dt^ix_i
$$ 
be in $\ker\phi$. Set 
$$
y_{d-1}:=x_d,
$$ 
and
$$
y_i:=By_{i+1}+x_{i+1}\quad\text{for}\quad i=d-2,d-3,\dots,0.
$$ 
Then it is straightforward to check, using our assumption 
$$
\sum_{i=0}^dB^ix_i=0,
$$ 
that we have 
$$
(tI-B)\ \sum_{i=0}^{d-1}t^iy_i=\sum_{i=0}^dt^ix_i,
$$ 
as desired.
EDIT. Here is a variant: The equality
$$
\sum_{i=0}^dB^ix_i=0,
$$ 
implies
$$
\sum_{i=0}^dt^ix_i=\sum_{i=0}^d\ (t^iI-B^i)\ x_i=(tI-B)\ \sum_{i=0}^d\ \sum_{j=0}^{i-1}\ t^jB^{i-1-j}x_i.
$$
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