Let $R=\mathbb{C}[x,y,z]/(x^2-yz)$ and $I=(\overline{x},\overline{y}) \subset R$. Prove that $I \cong \mathrm{Hom}_R(I,R)$ as $R$-modules.

I find $R \cong \mathbb{C}[st,s,st^2] \subset \mathbb{C}[s,t]$, $I \cong (st,s)$, and probably this is useful. Is there any good way? Thanks!


1 Answer 1


$\mathbb C[x,y,z]$ is a UFD and $y$ is coprime with $x^2-yz$, so $\bar y$ is not a zero divisor in $R$. Therefore multiplication by $\bar y$ gives an $R$-module isomorphism $\lambda:R\rightarrow\bar yR$. Also $\bar x^2=\bar y\bar z$, so $I^2\subseteq\bar yR$. Therefore we can define an $R$-module homomorphism $\phi:I\rightarrow\mathrm{Hom}_R(I,R)$ by $$ \phi(a)(b)=\lambda^{-1}(ab) $$ for $a,b\in I$. Note that $\phi(a)(\bar y)=\lambda^{-1}(a\bar y)=a$, so $\phi$ is injective.

Suppose $f\in\mathrm{Hom}_R(I,R)$ and let $\bar a=f(\bar y)$. Then $$ \bar yf(\bar x)=f(\bar x\bar y)=\bar xf(\bar y)=\bar a\bar x. $$ Pick a representative $a\in\mathbb C[x,y,z]$ for $\bar a$. The above gives $$ ax\in(y,x^2-yz)=(y,x^2), $$ so $a\in(y,x)$. Hence $\bar a\in I$. Finally $$ f(\bar x)=\lambda^{-1}(\bar a\bar x)=\phi(\bar a)(\bar x), $$ $$ f(\bar y)=\bar a=\phi(\bar a)(\bar y), $$ so $f=\phi(\bar a)$. Hence $\phi$ is bijective.


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