3
$\begingroup$

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!

$\endgroup$
0

1 Answer 1

6
$\begingroup$

$\mathbb C[x,y,z]$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.