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Let $K$ be a field (not algebraically closed) and $F$ be its algebraic closure. Let $X \subseteq K^n$ be Zariski closed, and $Y$ be the Zariski closure of $X$ inside $F^n$. Is it true that $\dim(X) = \dim(Y)$?

By $\dim(Z)$ I mean the Krull dimension of the Noetherian topological space $Z$ (maximum length of a chain of irreducible subsets of $Z$).

It is easy to see that $\dim(X) \leq \dim(Y)$, and that, in general, a subset $Z$ and its closure can have different dimensions. If equality fails, I would like an example with $K = \mathbb R$; if instead equality is always true, I would like a reference (but of course proofs and discussions are welcome).

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  • $\begingroup$ Do you mean something like $x^2 + y^2 + z^2 = 0$ in $\mathbf R^3$, which has only one $\mathbf R$-point, but over $\mathbf C$ becomes a surface? Or do you mean dimension in the scheme-theoretic sense, regardless of whether $X$ has enough points defined over $K$? $\endgroup$ Commented Sep 27, 2024 at 16:12
  • $\begingroup$ I mean neither. Treat $K^n$ and $F^n$ as topological spaces (ignore completely the algebra and geometry). Thus, if $X$ is finite, its closure inside $F^n$ is equal to $X$ itself. Example: if $K = \mathbb R$, let $Z \subset \mathbb R^2$ be the graph of the exponential function. Then $\dim(Z) = 1$, but the dimension of the Zariskl closure of $Z$ is 2 (because the closure is all $\mathbb R^2$). $\endgroup$ Commented Sep 27, 2024 at 16:22
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    $\begingroup$ Ah I see, I didn't read it correctly. I'll leave my comment up in case someone else has the same confusion. $\endgroup$ Commented Sep 27, 2024 at 16:28
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    $\begingroup$ This holds for all fields that are “big” or “ample” in the sense of Florian Pop. In particular, it holds for the field of real numbers. $\endgroup$ Commented Sep 27, 2024 at 23:59
  • $\begingroup$ @JasonStarr can you give a reference, please? I may mention this fact in a paper $\endgroup$ Commented Sep 28, 2024 at 8:28

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This is not true for an arbitrary field $K$. It may be true for $K= \mathbb R$.

Let $K=\mathbb Q$ and let $A$ be a simple abelian surface over $\mathbb Q$ with positive Mordell-Weil rank. Then $A(K)$ is Zariski dense in $A(F)$ so $A(K)$ is a Zariski closed subset of $\mathbb P^n(\mathbb Q)$. Restricting to some affine hyperplane, we get a Zariski closed subset of $\mathbb A^n(\mathbb Q)$. However, every proper Zariski closed subset of $A(\mathbb Q)$ is finite: It is the set of $\mathbb Q$-points of some curve, which can't be genus $0$ because $A$ is an abelian variety, and can't be genus $1$ because $A$ is simple, so must have finitely many rational points by Faltings' theorem. So the Krull dimension of $X$ (which is $A(\mathbb Q)$ minus finitely many points) is $1$, not $2= \dim A$ which is the Krull dimension of the Zariski closure of $A$.

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  • $\begingroup$ Thank you for your answer. Since I'm not familiar with many of the terms you use, could you give an explicit form (i.e. the equations for the example)? E.g.: how large is n? $\endgroup$ Commented Sep 27, 2024 at 16:13
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    $\begingroup$ @AntongiulioFornasiero Let $f$ be a polynomial in one variable of degree $5$ over $\mathbb Q$. One can express $f(x)f(y)$ as a polynomial $g(s_1,s_2)$ in $s_1=x+y$ and $s_2=x+y$. Then the pair of equations $z^2 = g(s_1,s_2), w(s_1^2-4s_2)=1$ in four variables $s_1,s_2, z,w$ is expected to work for about half of polynomials $f$, and it is easy but not enlightening to find an explicit $f$ that works. Probably a 3 variable version can be done as well. $\endgroup$ Commented Sep 27, 2024 at 17:59
  • $\begingroup$ There's a misprint: maybe you meant $s_2 = x-y$? $\endgroup$ Commented Sep 27, 2024 at 18:16
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    $\begingroup$ @AntongiulioFornasiero Rather I meant $xy$, apologies. $\endgroup$ Commented Sep 27, 2024 at 18:17
  • $\begingroup$ If I understand correctly, one of the claims is that the set of triples $(s_1, t, z) \in L^3: z^2 = f(s_1 + \sqrt t)f(s_1 - \sqrt t) \wedge t \neq 0$ has dimension 1 if $L = \mathbb Q$, while it has dimension 2 if $L= \mathbb C$ (for many choices of $f$)... $\endgroup$ Commented Sep 27, 2024 at 18:25
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$\DeclareMathOperator\algdim{algdim}$Here is a proof of what Jason Starr claimed (at least, in characteristic 0).

Let $K$ be a large field (in the sense of [Pop]) of characteristic $0$ and $F$ be its algebraic closure. Let $X\subseteq K^n$; define the algebraic dimension of $X$ as the Krull dimension of the Zariski closure $Y$ of $X$ inside $F^n$ (see [Dries]). If $X$ is Zariski closed inside $K^n$, then $\dim(X)$, the Krull dimension of $X$ (as a Noetherian topological space), is equal to its algebraic dimension.

Proof. By [Pop], we can assume that $K \subseteq K[[t]] \subseteq K^*$, where $K^*$ is an ultrapower of $K$. The same equations that define $Y$ also define some set $X' \subseteq K[[t]]^n$ and $X^* \subseteq (K^*)^n$. Since $K^*$ is an ultrapower of $K$ we have that $\dim(X) = \dim(X^*)$ and $\algdim(X) = \algdim(X^*)$. Since both $\dim$ and $\algdim$ are increasing, it suffices to show that $\dim(X') \geq \algdim(X')$. Thus, w.l.o.g., we may assume that $K$ is a Henselian field of characteristic $0$. By [Dries] Corollary 3.13, the projection of $X$ on some $K^d$ contains an open set $U$ (where $d := \algdim(X)$). Moreover, $\dim(U) =d$ and therefore $\dim(X) \geq d$.

[Dries] Dries, Lou van den. 1989. “Dimension of Definable Sets, Algebraic Boundedness and Henselian Fields.” Annals of Pure and Applied Logic 45:189–209.

[Pop] Pop, Florian. 2014. “Little Survey on Large Fields - Old & New.” In Valuation Theory in Interaction, 432–63. Congress Reports. European Mathematical Society.

There is the issue if the condition that $K$ is large is necessary….

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    $\begingroup$ On the real the proof is even easier: one can work directly inside $\mathbb R$ with the Euclidean topology. If $algdim(X) = d$, then the projection of $X$ on some $\mathbb R^d$ contains an Euclidean open set, and therefore $dim(X) \geq d$. $\endgroup$ Commented Sep 28, 2024 at 13:06
  • $\begingroup$ When you say "There is the issue if the condition that $K$ is large is necessary…", maybe you mean whether it is an exact characterisation—and that would be a piece of luck!—but of course @WillSawin's answer shows that certainly some condition is necessary. $\endgroup$ Commented Sep 28, 2024 at 13:11
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    $\begingroup$ @LSpice yes I meant: is it true that a field (of char 0) is large if and only if, for every X Zariski closed subset of $K^n$, $dim(X) = algdim(X)$? A possible candidate for a counterexample could be : Johnson, Will, and Jinhe Ye. 2023. “Curve-Excluding Fields,” 1–30. arxiv.org/abs/2303.06063. $\endgroup$ Commented Sep 28, 2024 at 13:16
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    $\begingroup$ Thank you for writing out the details. $\endgroup$ Commented Sep 28, 2024 at 13:28
  • $\begingroup$ My mistake: all algebraically bounded fields (and in particular the in particular the curve-excluding fields) satisfy the condition. $\endgroup$ Commented Sep 29, 2024 at 9:43

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