1
$\begingroup$

I am reading Eisenbud and Harris's The Geometry of Schemes. Exercise I-20 in it is to calculate the points and sheaf of functions for some schemes.

$1)$ $X=$Spec $\mathbb C[x]/(x^{2}-x)$. We know that it only has two closed points: $(x),(x-1)$. Call them $a,b$. The topology then should be$\{\emptyset,\{a\},\{b\},\{a,b\}\}$. We have that $\mathcal O(\emptyset)=0$,$\mathcal O(\{a,b\})=\mathbb C[x]/(x^{2}-x)$. Now I want to calculate $\mathcal O(\{a\})$

Note that $\{a\}=\{(x)\}=X_{x-1}, X_{x-1}$ is the basic open subset of $X$. So $\mathcal O(\{a\})=\mathcal O(X_{x-1})=(\mathbb C[x]/(x^2-x))_{x-1}$.

Am I right? If I am right, how to simplify $(\mathbb C[x]/(x^2-x))_{x-1}$

$\endgroup$
  • $\begingroup$ Use that $x^2-x=x(x-1)$, so $(\mathbb{C}[x]/(x^2-x))_{x-1}=\mathbb{C}[x]/(x)\simeq\mathbb{C}$. $\endgroup$ – Roland Feb 22 at 9:39
  • $\begingroup$ @Roland Could you explain why $(\mathbb C[x]/(x^2-x))_{x-1}=\mathbb C[x]/(x)$? $\endgroup$ – Mike Feb 22 at 10:20
  • $\begingroup$ $(\mathbb{C}[x]/(x^2-x))_{x-1}=(\mathbb{C}[x]_{x-1})/(x^2-x)$. But since $x-1$ is invertible, we have the equality of ideals $(x^2-x)=(x)$ in $\mathbb{C}[x]_{x-1}$. $\endgroup$ – Roland Feb 22 at 10:23
  • $\begingroup$ I get it. Thank you! $\endgroup$ – Mike Feb 22 at 10:37

Your Answer

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

Browse other questions tagged or ask your own question.