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edited Sep 28, 2024 at 13:09

Here$\DeclareMathOperator\algdim{algdim}$Here is a proof of what Jason Starr claimedclaimed (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⊆K^n$$X\subseteq K^n$; define the algebraic dimension of $X$ as the Krull dimension of the Zariski closure $Y$ of $X$ inside $F^n$ (see [Dries]). If $X$ is ZarioskiZariski closed inside $K^n$, then $dim(X)$$\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^*)$$\dim(X) = \dim(X^*)$ and $algdim(X) = algdim(X^*)$$\algdim(X) = \algdim(X^*)$. Since both $dim$$\dim$ and $algdim$$\algdim$ are increasing, it suffices to show that $dim(X') \geq algdim(X')$$\dim(X') \geq \algdim(X')$. Thus, w.l.o.g., we may assume that $K$ is a Henselian field of characteristic $0$. By [Dries] Corollary 3.13, the projection of $X$ on some $K^d$ contains an open set $U$ (where $d := algdim(X)$$d := \algdim(X)$). Moreover, $dim(U) =d$$\dim(U) =d$ and therefore $\dim(X) \geq d$.

[Dries] Dries, Lou van den. 1989. “Dimension of Definable Sets, Algebraic Boundedness and Henselian Fields“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“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..necessary….

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 $X⊆K^n$; 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 Zarioski 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...

$\DeclareMathOperator\algdim{algdim}$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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answered Sep 28, 2024 at 13:03

Antongiulio Fornasiero

Here is a proof of what Jason Starr claimed (at least, in characteristic 0).

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...