# Understanding Hartshorne's example II 3.2.2

Example 3.2.2. If $$P$$ is a point of a variety $$V$$, with local ring $$\mathcal{O}_P$$, then $$X:=\operatorname{Spec} \mathcal{O}_P$$ is an integral noetherian scheme, which is not in general of finite type over $$k$$.

First question: What does finite type over $$k$$ mean?

Is it that there must be a morphism from $$k$$ to $$X$$?

Trying to understand the example: As $$V$$ is a variety, then $$A(V)$$ is a domain, so $$A(V)_{\mathcal{m}_P}\cong \mathcal{O}_P$$ (Theorem I.3.2 (c)) is a domain, so $$\mathcal{O}_P$$ is a domain, so $$X$$ is integral. As $$\mathcal{O}_P$$ is noetherian, then $$X$$ is noetherian.

There is a isomorphism $$A(V)\cong\mathcal{O}_P$$ (Theorem I.3.2 (a)) and a map from $$\mathcal{O}(V)\to \mathcal{O}_P$$, so we get a map $$A(V)\to \mathcal{O}_p$$. On the other hand there is another map $$k\to A(Y)$$ so composing these maps I get $$\varphi:k\to \mathcal{O}_P$$ Let $$X:= \operatorname{Spec} \mathcal{O}_P$$ and $$Y:\operatorname{Spec} k$$. The map $$f$$ induces another map $$f:X\to Y$$

But, $$\mathcal{O}_P$$ need not be necessarily a finitely generated $$k$$-algebra. So $$f$$ is not of finite type.

Is it ok?

Edited If it is ok, is there any example of a variety $$V$$ such that $$O_P$$ is not a finitely generated $$k-$$algebra?

Thank you.

• You can look up finite type practically anywhere, including in Hartshorne's book: the definition is in the middle of page 84. – KReiser 2 days ago
• @KReiser thank you for your answer, in the definitions it says: "A morphism $f:X\to Y$ is of finite type if ...". Should it be, "a morphism $f:X\to Y$ is a finite type over $X$ if..."? – Framate 2 days ago
• I mean, $f:X\to Y$ is a finite type over $Y$. – Framate 2 days ago
• No, that's not how those words are used. You can say "$f$ is of finite type" or "$X$ is of finite type over $Y$". – KReiser 2 days ago
• @KReiser Thanks a lot! – Framate 2 days ago