# Finding a presentation matrix for an $F[t]$-module

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.

• If $V$ is generated by the $v_i$s as an $F$-vector space, clearly, the same generators generate it as an $F[t]$-module. Thus, as you said, you have a surjection $F[t]^n\to V$. You should show that the kernel is also $F[t]^n$ and thus the map $F[t]^n\to F[t]^n$ is given by an $n\times n$ matrix over $F[t]$. You should show (for suitable basis for the $F[t]^n$), that this matrix is $tI-B$. – Mohan Apr 6 '18 at 22:49
• @Mohan Thanks, I understood your first two sentences (but unfortunately not the succeeding ones). However, I think I got what I'm asked to do (I stated it in the edited version of my question). What I'm confused by now is perhaps some trivial fact: by definition, $T(v_i)=tv_i$, so it seems like the matrix $B$ is the scalar matrix $tI$. But a matrix of a linear operator of $F$-vector spaces is supposed to have entries in $F$... Also if $B=tI$, then $A=0$. – Cary Apr 7 '18 at 1:43
• @Cary - Are you sure the exercise is stated as in your first block quote, and that $T$ is defined as you wrote? – Pierre-Yves Gaillard Apr 7 '18 at 15:32
• @Pierre-YvesGaillard I checked the statement of the exercise, and I believe it is correct. However the original exercise doesn't talk about the definition of $T$; since throughout the corresponding chapter (14.8) $T$ stood for the operator I defined, I assumed the definition is the same in the exercise too. – Cary Apr 7 '18 at 16:06
• There is a solution here (p. 68, exercise 14.8.4 csie.ntu.edu.tw/~b01902113/artin-sols.pdf), but I don't understand its solution mainly because I don't understand what exactly $B$ is and what the meaning of things like $Bv_i$ is ($B$ is a matrix, and it can only be multiplied by the coordinate vector of $v_i$, but it cannot be applied to $v_i$). – Cary Apr 7 '18 at 16:10

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 $$\sum_{i=0}^dt^ix_i=\sum_{i=0}^d\ (t^iI-B^i)\ x_i=(tI-B)\ \sum_{i=0}^d\ \sum_{j=0}^{i-1}\ t^jB^{i-1-j}x_i.$$
• Thanks! I'm not sure I understand what $\phi$ is (the formula is unclear since elements of the domain $R^n$ are column vectors with entries in $F[t]$, but I don't know what $x_i$s are). According to your verbal description, there is a map $F^n\to V$ given by $X\mapsto \mathbf BX$ with inverse $V\to F^n$ given by $v_1x_1+\dots v_nx_n\mapsto (x_1,\dots,x_n)^t$. There is also a map $R^n\to V$ defined in my question. So I guess your $\phi$ should be the composition of the latter map $R^n\to V$ with the former map $V\to F^n$. But I don't see why your formula defines this map. – Cary Apr 7 '18 at 19:40
• And also I have another concern: if you show that $(*)$ is exact, then you will show that $A$ is a presentation matrix for $F^n$, not for $V$. Of course they are isomorphic as $F$-vector spaces, but why do the corresponding $F[t]$-modules have the same presentation matrix? – Cary Apr 7 '18 at 19:46
• @Cary - First comment: $$R^n\ni u=(u_1,...,u_n),\ u_j\in R,\ u_j=\sum_iu_{ji}t^i,\ u_{ji}\in F,$$ $$x_i:=(u_{1i},...,u_{ni})\in F^n,\ t^ix_i:=(u_1t^i,...,u_nt^i)\in R^n,$$ $$u=\sum t^ix_i.$$ – Pierre-Yves Gaillard Apr 7 '18 at 19:57
• @Cary - Comment 2: If $R^n\xrightarrow{tI-B}R^n\xrightarrow\phi F^n\to0$ is exact, and if $\psi:F^n\to V$ is an isomorphism, then $R^n\xrightarrow{tI-B}R^n\xrightarrow{\psi\circ\phi}V\to0$ is exact (because $\psi\circ\phi$ is surjective and has the same kernel as $\phi$). – Pierre-Yves Gaillard Apr 7 '18 at 20:13