Can a discrete valuation ring be finitely generated over a field?

Question

In my homework of schemes, the professor proposed the following exercise:

"Let $ X $ be a scheme of finite type over a field $ k $ and $ f \in \mathcal{O}_X (X) $ a global section. Show that $ f $ is nilpotent $\iff f (x) = 0 $ for every $ x $ closed point in $ X $. Show by a counterexample that this assertion is false in general, that is, if we no longer assume $ X $ to be a scheme of finite type over a field."

Before proving the main assertion, I will exhibit a contre-example based of the spectrum of a Discrete Valuation Ring:

(Counterexample when $ X $ is not a scheme of finite type over a field):

Let $ X = Spec A $ with $ (A, m) $ a discrete valuation ring. Topologically $ X = \{x, \eta \}$ where the open sets are $ \{\emptyset, X, \{\eta \} \} $ and the closed ones are $ \{\emptyset, X, \{x \} \} $. Let $ \pi $ be a uniformizing element of $ A $, that is, a generator of the maximal ideal $ m$. We know that $ \pi $ is not nilpotent, but $ \pi (x) = 0 $ where $ x $ is the unique closed point of $X$.

We can even find examples where $A$ a $k$-algebra. This is the case, for example of $A = k [[T]] $. The problem here will be precisely the hypothesis of finite type which is not respected:

Indeed if $ X = Spec k [[T]] $ and $ f = \pi $, Proposition 2 does not apply. More precisely, let $ U = D (f) \cong Spec \; \kappa (\eta) $, then: $ \eta $ is a closed point on $ U $, whose residual field is $\kappa (\eta) $ is a transcendent extension of $k$, and which is not closed on $X$. The problem is that $ \mathcal{O}_U (U) $ and $\mathcal{O}_X (X) $ are not finitely generated $k$-algebras.

(My question): "Assuming that the main assertion of the exercise is true and that my contre-example based the spectrum of a DVR is right, I have the impression that this shows indirectly that a discrete valuation ring cannot be a finitely generated $k$-algebra.

It is evident for the case when $A = k [[T]]$, but is there a more direct proof of this apparently true fact?

If so, does this result have any relation with the fact that DVR over a field $k$ are usually stalks of the sheafs of regular functions in non-singular points of algebraic curves over $k$?

(Proof of the main assertion)

We start by recalling some properties of the nilpotent global sections of quasi-compact schemes:

(Proposition 1): Let $Z$ be a quasi-compact scheme. So for all $f \in \mathcal{O}_Z (Z) $, we have: $f$ vanishes on all points of $Z \iff f $ is nilpotent.

(Proof - Prop 1):

1. $ (\Rightarrow) $ If $ Z $ is affine, it is straightforward from: $ \bigcap \limits_ {p \subset B} \, p = \sqrt {( 0)} $.

If $ Z $ is not affine, we take a finite cover $ Z = \cup_i U_i $ by affine open subschemes. Since the proposition is true for affine schemes, $ f_ {| U_i} $ is nilpotent for all $ i $.

By hypothesis, we only have a finite number of $ i $, so there exists a $ N $ such that: $ f_ {| U_i} ^ N = 0 $, $ \forall i $. Since $\mathcal{O}_Z$ is a sheaf, the $ f_ {| U_i} ^ N $ glue to $ f ^ N $, which is therefore, equal to $ 0 $.

2. $(\Leftarrow)$ It's trivial.

In order to study the nilpotent global sections of a scheme of finite type over $k$. We first recall the following property:

(Proposition 2) Let $ X $ be a scheme of finite type over a field $ k $. For any point $ x \in X $, the following assertions are equivalent:

1. $ x \in X $ is a closed point.
2. $ k \hookrightarrow \kappa (x) $ is a finite extension.
3. $ k \hookrightarrow \kappa (x) $ is an algebraic extension.

(Proof - Prop 2) Let $ U = Spec \, A $ be an affine open subscheme containing $ x $. In $ U $, the point $ x $ corresponds to a prime ideal $ p \subset A $. We then consider the compositions $ k \hookrightarrow A \to \frac{A}{p} \hookrightarrow Frac (\frac {A}{p}) = \kappa (x) $.

• $(1 \Rightarrow 2) $: In this case, $x$ is also closed in $ U $ and $p $ is therefore a maximal ideal of $ A $. This implies that $ \kappa (x) = \frac {A} {p} $ and, it is both a field and a $ k $-algebra of finite type . According to the Nullstellesatz, $ k \hookrightarrow \kappa (x) $ is therefore a finite extension.

• $ (2 \Rightarrow 3) $: Any finite field extension is algebraic.

• $ (3 \Rightarrow 1) $: As $\kappa (x) / k $ is assumed to be algebraic, the composition $ k \ to \frac {A} {p} $ is an integral morphism. Since it is injective, it can be identified with an integral extension of integral domains. In this case: $ k $ is a field $\Rightarrow \frac {A} {p} $ is a field and, this shows that $ x $ is a closed point of $ U $ .

  We have just shown that if $ \kappa (x) / k $ is algebraic, then: for every affine open subset $ U \ni x $, $ x $ is a closed point of $ U $. It is a standard fact that a point is closed in a scheme if and only if it's closed in every affine open subscheme, hence $ x $ is a closed point of $ X $.

(Solution to the exercise):

Since $ X $ is quasi-compact, Proposition 1 shows that $ f \in \mathcal{O}_X (X) $ is nilpotent $ \iff $ $ D (f) = \emptyset $. In general, $ D (f) $ is a scheme of finite type over $ k $. Since $ D (f) $ is in particular Noetherian, $ D (f) = \emptyset \iff D (f) $ has no closed point.

By Proposition 2: $ D (f) $ has no closed point $ \iff $ No point in $D(f)$ is closed in $X \iff f (x) = 0 $, $ \forall x $ closed point in $ X $.

Answer 1

You're correct: a discrete valuation ring cannot be finitely generated over $k$. Suppose $R$ was finitely generated over $k$. Then $R[t]$ would be as well, and so would $R[t]/(t\pi-1)$. But this last ring is the field of fractions of $R$, which is of transcendence degree one over $k$, contradicting Zariski's lemma (your proposition 2).