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.

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    $\begingroup$ You can look up finite type practically anywhere, including in Hartshorne's book: the definition is in the middle of page 84. $\endgroup$ – KReiser 2 days ago
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    $\begingroup$ @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..."? $\endgroup$ – Framate 2 days ago
  • $\begingroup$ I mean, $f:X\to Y$ is a finite type over $Y$. $\endgroup$ – Framate 2 days ago
  • $\begingroup$ No, that's not how those words are used. You can say "$f$ is of finite type" or "$X$ is of finite type over $Y$". $\endgroup$ – KReiser 2 days ago
  • $\begingroup$ @KReiser Thanks a lot! $\endgroup$ – Framate 2 days ago

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