Understanding Hartshorne's example II 3.2.2

Question

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.

Answer 1

Here is an explanation for giving an explicit counterexample. Thinking in terms of affine varieties, we are looking for $V =\mathrm{Spec} A$ and a maximal ideal $\mathfrak{m}$ in $A$ such that:

1. $A$ is a finite type $k$-algebra;
2. the localization of $A$ at $\mathfrak{m}$ is not of finite type as a $k$-algebra.

I'll work in characteristic $0$ for simplicity's sake. If $k$ is an uncountable field of characteristic $0$ (think $k=\mathbb{R}$ or $\mathbb{C}$), then you can consider $A = k[X]$, which is of obviously of finite type over $k$. The corresponding variety is the affine line $V:= \mathbb{A}_k^1$.

Localization of $A$ at $(X)$ gives you the local ring at the origin $k[X]_{(X)}$ of $k[X]$, i.e. take $P = (0)$, then $\mathcal{O}_P = k[X]_{(X)}$.

Now $k[X]_{(X)}$ is not of finite type over $k$, as pointed out in the second answer here. A way of seeing this is by looking at he dimension as a $k$-vector space: $\dim_k k[X]_{(X)}$ is uncountable, as it admits a free family $\left( \frac{1}{X-a} \right)_{a\in k^{\times}}$. On the other hand any finitely generated $k$-algebra has at most countable dimension over $k$ (being a quotient of a polynomial ring).

Thus $\mathrm{Spec} (k[X])$ is a variety over $k$, but $\mathrm{Spec} (\mathcal{O}_P )$ is not of finite type over $k$.