# 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

Staff Emeritus
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

Staff Emeritus
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].

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