What is the Presentation and Determination of Modules over a Field?

[Kreizhn]

Homework Statement

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.

Homework 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.

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?

 

[micromass]

Homework Statement

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.

Homework 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.

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?

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.

 

[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.

 

[micromass]

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.

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].