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Let $S$ be a Dedekind scheme with function field $K=K(S)$ and $C$ a projective regular curve over $K$, so we can fix certain closed embedding $e:C \subset \mathbb{P}^n_K$.

Let compose this embedding with canonical map $f: \mathbb{P}^n_K \to \mathbb{P}^n_S$ and consider the Zariski closure $Z_C:=\overline{f\circ e(C)}$ in $\mathbb{P}^n_S$ of the image of $C$.

Q ( the "initial" one): Is it true that $Z$ has dimension $2$ and if yes, how to see it?
(#Edit: Think so, if the sketched argument in "Idea" below works)

Can the statement be generalized (#1) to higher dimensions in following way: Say $X \subset \mathbb{P}^n_K$ is an irreducible subscheme of demension $d$ and we take as before the closure $Z_X:=\overline{f\circ e(X)}$ of the image of $X$ in $\mathbb{P}^n_S$. What can we say about it's dimension?

So the point becomes if that's a specific curve-to-surface feature, or does it hold in higher dimensions as well?

an "idea"/ plagiarism: If I'm not missing something the same argument as Daniel Loughran gave here should go through in this problem as well, should't it?

Namely, the induced projection map $p:Z_X \to S$ is dominant, so flat, since we are over Dedekind domain $S$ and by assumption the generic fiber has dimension $d$ of $X$, so by this flatness argument $ Z_X$ should have dimension $d+1$ if we add up fiber and base dimensions (which works for dominant flat maps since flatness assures that the fiber dimension stays constant). Is the exposed argumentation correct so far?

If the sketched argument in the Idea above works, then this leads me to the "natural" attempt to generalize this this question once more by now dropping the flatness feature, which as $S$ Dedekind, was in above formulations "donated for free" to us :

Generalisation #2: What can we say about the dimension of the closure $Z_X \subset \mathbb{P}^n_S$ if $S$ would be instead an arbitrary irreducible locally Noetherian scheme with field of fractions $K$? (Note that induced map $Z_X \to S$ would be still dominant, so surjective due to properness.

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I'll interpret your terminology "Dedekind scheme" to mean "regular integral locally Noetherian scheme of dimension one" (or dimension $\leq 1$ if you replace $d + 1$ with $d + \operatorname{dim} S$ in your notation).

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When $S$ is a Dedekind scheme, the answer to your question (for $Z$ of arbitrary dimension) is yes via flatness as you argue (e.g.Tag 0D4J and Tag 0AFE). (Small comment: for flatness of $Z_X \rightarrow S$, being dominant is not enough, e.g. there could be an embedded point of $Z_X$ lying over a closed point of $S$ like $\operatorname{Spec} k[x,y]/(xy, y^2) \rightarrow \operatorname{Spec} k[x]$ for a field $k$. But it will be ok in your situation, where you are taking scheme-theoretic closure.)

Remark: The properness is being used here (but projectivity is not important). Suppose $S = \operatorname{Spec} \mathbb{Z}_{(p)}$ and replace $\mathbb{P}^n_S$ in your notation with $\mathbb{A}^2_S = \operatorname{Spec} \mathbb{Z}_{(p)}[x,y]$. Consider the curve $C = \operatorname{Spec} \mathbb{Q}[x,y]/(p x - 1)$ in the generic fiber. The closure of $C$ in $\mathbb{A}^2_S$ is $\operatorname{Spec} \mathbb{Z}_{(p)}[x,y]/(px - 1)$. But this scheme has empty fiber over the closed point of $S$.

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Next for your Generalisation #2. Probably we should at least assume $S$ is finite dimensional (as $S$ could have infinite dimension like in Tag 02JC). I will also assume $S$ is reduced. (This is harmless because the reduced closed subscheme $S_{\mathrm{red}} \rightarrow S$ is a universal homeomorphism, and since scheme-theoretic closure along quasi-compact morphisms has underlying set coinciding with set-theoretic closure).

I will add some some more hypotheses: assume $S$ is universally catenary, Jacobson, and admits a (necessarily unique) "dimension function" $\delta \colon S \rightarrow \mathbb{Z}$ which sends closed points to $0$ (as in Tag 02QK). This dimension function $\delta$ sends any point to the (topological) dimension of its closure.

These additional hypotheses on $S$ are satisfied, for example, if (a) $S$ is locally finite type over a field or if (b) $S$ is locally of finite type over a Dedekind domain with infinitely many primes.

Then Tag 02QO implies that what you ask for indeed holds. More precisely/generally: Suppose $X$ is a scheme which is locally of finite type over $S$. Suppose $Z \subseteq X_K$ is an irreducible closed subscheme of the generic fiber, with closure $\overline{Z} \subseteq X$. Then $\operatorname{dim} \overline{Z} = \operatorname{dim} Z + \operatorname{dim} S$.

Remark: Note that there are no properness hypotheses in the result above. This result is false in general if $S$ is not Jacobson (e.g. $S = \operatorname{Spec} \mathbb{Z}_{(p)}$ as in the remark above).

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I don't have an answer for your Generalisation #2 outside the cases covered above.

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  • $\begingroup$ about the second paragraph: as you mentioned the presence of embedded primes in $Z_X$ prevents $Z_X \to S$ for $S$ Dedekind from beeing flat, because these give zero divisors und this leads to existence of torsion, so indeed no flatness between stalks at such points. But then you wrote that in above situation it's ok when $Z_X$ is formed as scheme theoretic closure of $X$ inside $\mathbb{P}^n_S$. Why? By definition the schematic closure $Z_X$ of $X$ with resp the immersion $i:=f \circ e$ of $X$ is given as closed subscheme of $\mathbb{P}^n_S$ with resp the largest qcoh ideal sheaf $\endgroup$ Commented Oct 11, 2023 at 9:22
  • $\begingroup$ let's call it $\mathcal{I}'$ - contained in $ \mathop{\mathrm{Ker}}(\mathcal{O}_{\mathbb{P}^n_S} \to f_*\mathcal{O}_ X)$. ( note the latter might be not qcoh, thats why we need $I'$) But the question is why such one $V(\mathcal{I}')$ cannot have embedded primes, and so forming scheme theoretic closure avoids these problems as your last sentence in second paragraph suggests? For example, I not see any reason why it should be reduced (which would indeed exclude these troubles with embedded points preventing the map from beeing flat as you exposed above) $\endgroup$ Commented Oct 11, 2023 at 10:30
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    $\begingroup$ @user267839 If $X$ (in your notation) was reduced to begin with, then the scheme-theoretic closure is also reduced Tag 056B. This is good enough for your question about dimension, which is just about the underlying topological spaces. By the way, your $\mathop{\mathrm{Ker}}(\mathcal{O}_{\mathbb{P}^n_S} \to f_*\mathcal{O}_ X)$ is in fact quasi-coherent because $X \rightarrow \mathbb{P}^n_S$ is a quasi-compact morphism by the locally Noetherian hypotheses (then apply Tag 01R8). $\endgroup$ Commented Oct 11, 2023 at 14:25
  • $\begingroup$ @user267839 Even in the non-reduced case, we can argue as follows. Note that $Z_X$ is the scheme-theoretic closure of its generic fiber. Working affine-locally on $Z_X$, we then see that flatness of $Z_X \rightarrow S$ reduces to the following commutative algebra fact: if $A$ is a Dedekind domain with fraction field $K$ and $B$ is an $A$-algebra such that $B \rightarrow B \otimes_A K$ is injective then $B$ is flat over $A$ (it's clearly torsion free). $\endgroup$ Commented Oct 11, 2023 at 14:39
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Let $f: X \rightarrow Y$ be a proper dominant morphism between two locally Noetherian integral schemes.

By Stacks Lemma 02JX, we have $\dim{X}=\dim{Y}+\delta$, where $\delta$ is the transcendence degree of the field of functions of $X$ over that of $Y$.

But it’s classical that for any finitely generated $k$-algebra $A$ which is a domain, the Krull dimension of $A$ is exactly the transcendence degree of $Frac(A)$ over $k$. So $\delta$ is the dimension of the generic fibre.

Thus, if $\eta \in Y$ is the generic point, $\dim{X}=\dim{Y}+\dim{X_{\eta}}$.

Of course, this still works when we only assume that $X,Y$ are irreducible instead of integral, and it shouldn’t be hard to generalize to the case where $X \subset \mathbb{P}^n_Y$ is defined as the closure of some closed subscheme of $\mathbb{P}^n_{\eta}$ (closure commutes with finite unions, and dimension is well-behaved with respect to finite unions of closed subsets).

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  • $\begingroup$ about the part where we can assume that $X,Y$ that irreducible instead of integral: It seems that the desired formula $\dim{X}=\dim{Y}+\dim{X_{\eta}}$ follows immediately from integral case, since passing to reduced structures $X\to X_{\text{red}}$ not affect the dimensions. The only term that possibly could potentially cause troubles is the generic fiber. Do you see an ad hoc reason why $(X_{\text{red}})_{\eta}$ should have same dimension as $(X_{\eta})_{\text{red}}$? $\endgroup$ Commented Oct 11, 2023 at 8:50
  • $\begingroup$ In more broad sense the last question is about the compatibility of dimension behavior for closed immersions $V(I) \subset X$ with respect the formation of fibers. If I'm not missing something then above the situation rather easy because the fiber of $\eta$ is dense in $X$, but let try to generalize it: Say $y \in Y$ is arbitrary and $V(I) \subset X$ a closed immersion for some qcoh ideal sheaf $I$, is there any interesting "controllable" relation between dimensions of $V(I)_y$ and $V(\overline{I}) \subset X_y$ where $\overline{I}$ is the qcoh ideal in $X_y$ generated by image of $I$ with $\endgroup$ Commented Oct 11, 2023 at 9:01
  • $\begingroup$ respect the can map $X_y \to X$, which is a priori just an immersion ( ie not a closed one as long as $y$ is choosen arbitrary)? $\endgroup$ Commented Oct 11, 2023 at 9:02
  • $\begingroup$ Remark: in the special situation above where $I$ is the nilporent ideal the story should be easy since passing to reduced scheme is a homeomorphism, so nothing "strange" could happen with fibers, and the generic fiber term cause no undesired dimension jumps when we pass to reduced structures as you suggest in last paragraph. But what about the more general situation with $I$ any quasi coherent ideal sheaf? $\endgroup$ Commented Oct 11, 2023 at 10:40
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    $\begingroup$ For the question in your first comment: that’s “just” because we’re always considering the same topological space. Re your third comment, $X_y \rightarrow X$ has no reason to be an immersion any more – say $X=Y$, then $\{y\} \rightarrow Y$ is an immersion iff $y$ is a closed point of some open subset of $Y$. So if $Y$ is of finite type over a field, this is equivalent to $y$ being a closed point. For your second comment, if $I$ is a qcoh ideal sheaf on a $Y$-scheme $X$, I would certainly expect $V(I)_y$ and $V(I_y)$ to be the same… isn’t it clear in the affine case? $\endgroup$ Commented Oct 11, 2023 at 11:37

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