Homework Help: Presentation of modules

1. Jul 13, 2011

Kreizhn

1. The problem statement, all variables and given/known data
Let k be a field and k[x] be the set of polynomials over that field. Given that M is a module with presentation
$$\begin{pmatrix} 1+ 3x & 2x & 3x \\ 1 + 2x & 1+ 2x -x^2 & 2x \\ x & x^2 & x \end{pmatrix}$$
determine M.

2. Relevant equations
One can apply elementary row and column operations. In the event that one reduces a row or column to the point that there is only one non-zero element and that element is a unit, we can remove the row and column of that unit and the corresponding presentation is isomorphic to the original one.

3. The attempt at a solution
So what we have here originally is a exact sequence
$$k[x]^3 \xrightarrow{\phi} k[x]^3 \to M \to 0$$
so that we may take $M \cong \text{coker}\phi$.

If I play around with the matrix a bit, I can reduce it to (x) modulo mistakes in my matrix manipulation. Thus we get a new homomorphism
$$k[x] \xrightarrow{\varphi} k[x] \to \tilde M \to 0$$
with $\tilde M \cong M$.

So here's the part I'm not too sure about. We have $\varphi: k[x] \to k[x]$, so does this mean that $1 \mapsto x$?

2. Jul 13, 2011

micromass

Yes. In fact, we have here that $$M\cong k[X]/\xi(k[X])$$, where $\xi$ is the map associated with the matrix. So, we have $$M\cong k[X]/(X)\cong k$$ here.

3. Jul 13, 2011

Kreizhn

Thanks micromass.

Yeah, that's what I had originally. Though I started doubting myself as to whether the image of the homomorphism was really just the ideal generated by x.

4. Jul 13, 2011

micromass

No need to doubt yourself The image of a k[X]-module is always a k[X]-module. Thus you can expect the image of k[X] to be a submodule of k[X], and submodules of k[X] are exactly the ideals of k[X].