Let $R$ be an integral domain, and consider the $R$-module map $\phi \colon R^2\to R$ described by the matrix $\phi = (a,b)$ for some $a,b\in R$, meaning $$ \begin{pmatrix}a&b\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = ax+by $$ My question is what the kernel is, or more precisely, if the kernel is $\ker\phi = (-b,a)^t R$. If not, under which circumstances is the kernel this submodule?
My question actually appears in the calculation of a free resolution of $\Bbb Z$ for the abelian group $G = \Bbb Z^2$. Hence, in my case, $R = \Bbb Z[G]\cong\Bbb Z[x,y]$, which is a polynomial ring over $\Bbb Z$ in two commuting variables. Moreover, I actually have $a = e_{(0,0)} - e_{(1,0)}$ and $b = e_{(0,0)} - e_{(0,1)}$ where $e_{(i,j)}$ denotes the $\Bbb Z$-basis element in $R$ associated to the group element $(i,j)\in\Bbb Z^2$.
