$\mathrm{Hom}_R (R/(x),R)\cong M$ for $R=k[x,y]/(xy)$. $M=R/(x)$

Question

Let $k$ be a field. Let $R=k[x,y]/(xy)$. $M=R/(x)$. Show that $M^{\vee}$:=$\mathrm{Hom}_R (R/(x),R)\cong M$. I know that $\mathrm{Hom}_R(R,M)\cong M$ in general for any left $R$-module $M$, but is it true that the in the above setting, $\mathrm{Hom}_R (R/(x),R)\cong M$?

Answer 1

Let $R$ be a ring, and fix and ideal $I\subset R$; denote the quotient map as $p:R\to R/I$.

For any $R$-module $M$, the precomposition with $p$ yields the following $R$-linear map: $$\rho:\mathrm{hom}_R(R/I,M)\to\mathrm{hom}_R(R,M).$$ Since $p$ is surjective, $\rho$ is injective; plus a $R$-linear map $f:R\to M$ factors through $p$ if and only if $f(I)=0$, so $\operatorname{im}\rho$ consists of the the $R$-linear maps vanishing on $I$.

Hence $\mathrm{hom}_R(R/I,M)$ is isomorphic via $\rho$ to the submodule $S\subset\mathrm{hom}_R(R,M)$ of the $R$-linear maps $f:R\to M$ with $f(I)=0$.

The isomorphism $\phi:\mathrm{hom}_R(R,M)\to M$, being defined as $\phi(f)=f(1)$, maps $S$ to the submodule $\{m\in M:I\cdot m=0\}$: in fact $f(r)=0$ iff $r\cdot \phi(f)=r\cdot f(1)=0$, and any $m$ is equal to $\phi(f)$ for a unique $f$.

Now set $R:=k[X,Y]/(XY)$, $M:=R/(x)$, where $x$ (and $y$) is the residue of $X$ (and $Y$) in $R$. As seen, $\mathrm{hom}_R(M,R)$ is isomorphic to the ideal $V:=\{r\in R:xr=0\}$, because $(x)r=0$ iff $xr=0$. Any $r$ can be written as: $a_nx^n+\dots+a_1x+b_my^m+\dots +b_1y+c$ with $a_i,b_j,c\in k$, so $xr=0$ if and only if $a_nx^{n+1}+\dots +a_1x^2+cx=0$, i.e. $r\in (y)$.

Thus $V=(y)$, and it is left to find a $R$-linear isomorphism $R/(x)\to (y)$. By the Noether theorems: $$R/(x)= (k[X,Y]/(XY))/((X)/(XY))$$ $$\cong k[X,Y]/(X)\cong k[Y].$$ So it suffices a isomorphism $\mu:k[Y]\to (y)$.

Define $\mu$ as: $$a_nY^n+\dots +a_1Y+a_0\mapsto a_ny^{n+1}+\dots +a_1y^2+a_0y.$$ This map is clearly $R$-linear, and it is injective since a polynomial of the form $a_nY^{n+1}+\dots +a_1Y^2+a_0Y$ is in $(XY)$ only if $a_n,\dots a_0=0$. Plus it is surjective, as every element of $(y)$ is of the form $a_ny^{n+1}+\dots +a_1y^2+a_0y$.