# Describe the points and the sheaf of functions of some schemes.

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 $$(\mathbb C[x]/(x^2-x))_{x-1}$$

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