I am trying to solve part (c) the following problem from Artin. I have already solved (a) and (b).
Determine whether or not the modules over $R=\mathbb{C}[x,y]$ presented by the following matrices are free:
(a) $\begin{bmatrix} x^2+1 & x \\ x^2y+x+y & xy+1 \end{bmatrix}$
(b) $\begin{bmatrix} xy-1 \\ x^2-y^2 \\ y \end{bmatrix}$
(c) $\begin{bmatrix} x-1 & x \\ y & y+1 \\ x & y \\ x^2 & 2y \end{bmatrix}$
I have determined that the matrix in (a) reduces to $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, and hence the module is isomorphic to $R^2/\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} R^2\cong R^2/R^2 \cong \{0\}$, hence not free. (According to my professor's definition, "free" means isomorphic to $R^k$ for some $k\in\{1,2,3,...\}$)
Likewise, I have determined that the matrix in (b) reduces to $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, and hence the module is isomorphic to $R^3/\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}R \cong R^2$, hence free.
Now, I have determined that the matrix in (c) reduces to $\begin{bmatrix} 1+2y \\ x+y \\ x^3+x-1 \end{bmatrix}$, but I'm having trouble reducing this matrix further. How can I do this?
Assuming your reduction to $$\begin{bmatrix}1+2y \\ x+y \\x^3+x-1\end{bmatrix}$$ is correct (I haven't checked), you may subtract the first line $(1/2)$-times from the second. That brings you to $$\begin{bmatrix}1+2y \\ x-\frac{1}{2} \\x^3+x-1\end{bmatrix},$$ where $ x-\frac{1}{2},\;\; x^3+x-1 \in \mathbb{C}[x],$ which is a PID. Note that the two polynomials are also coprime since $(1/2)$ is not a root of $x^3+x-1$. Thus, there are some polynomials $f(x), g(x)$ (that are computable, but that's not necessary) such that $$(x-\frac{1}{2})f(x)+(x^3+x-1)g(x)=1. \;\;\;\;(*)$$ That is, we have an invertible matrix $$A=\begin{bmatrix}1 & 0 & 0 \\ 0 & f(x) & g(x) \\ 0 & -(x^3+x-1) & x-\frac{1}{2} \end{bmatrix}$$ (note that the determinant is $1$ because of $(*)$), and $$A \cdot \begin{bmatrix}1+2y \\ x-\frac{1}{2} \\x^3+x-1\end{bmatrix}=\begin{bmatrix}1+2y \\ 1 \\0\end{bmatrix}.$$ Now, using the $1$ in the middle one easily eliminates the first entry. Hence, the module is free.