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Let $R$ be a commutative ring with unity and let $M$ be a finitely generated noetherian $R$-module.

Suppose we are given an $R$-homomorphism: $$ \varphi: R^{(2)}\to R^{(2)} $$ Fixing a basis of $R^{(2)}$ we can obtain $\varphi$ as a $2\times 2$ Matrix with entries in $R$. Let's say it's given as \begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix} This induces the map $$ \varphi_*:hom_R(R^{(2)},M)\to hom_R(R^{(2)},M),\quad f\mapsto f\circ\varphi $$ Now since $hom_R(R^{(2)},M)\simeq M\oplus M$, there should be a map "belonging" to $\varphi_*$ that goes from $M\oplus M$ to $M\oplus M$.

I tried to understand what this map exactly does and following the isomorphism $hom_R(R^{(2)},M)\simeq M\oplus M$ I came to the result that it somehow looks like this: $$ \varphi_*^{'}:M\oplus M \to M\oplus M,\quad (m_1,m_2)\mapsto (m_1,m_2)\left[\begin{array}{l}r_{11}&r_{12}\\r_{21}&r_{22}\end{array}\right]=(m_1r_{11}+m_2r_{21}, m_1r_{12}+m_2r_{22}) $$ Is this correct? And if yes how can I express the Image of that map, because I am having a hard time trying to do this.

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  • $\begingroup$ Yes, that's correct. It's hard to say much about the image of the map without more information. $\endgroup$ – Qiaochu Yuan Nov 27 '17 at 18:38
  • $\begingroup$ @QiaochuYuan ok, great. So in my case $R:=k[[x,y]]/(f)$ a quotient of the formal power series ring in two variables with $f=x^2+y^{n+1}$, $n$ an odd integer, $k$ an algebraically closed field, $char(k)=0$, $\varphi_j=\left[\begin{array}{l}x&y^j\\y^{n+1-j}&-x\end{array}\right]$ with $1\leq j\leq (n-1)/2$ and $M$ is actually the cokernel of $\varphi$. So yeah very confusing to me but I thought maybe there is just a simple way of thinking about the image of that map. $\endgroup$ – Jules Nov 27 '17 at 22:47

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