# Fiber Dimension Theorem for infinite-dimensional schemes

For varieties $X,Y$ over an algebraically closed field, and a surjective morphism $f:X\rightarrow Y$, $\dim f^{-1}(y)\geq\dim X-\dim Y$ for all closed $y\in Y$, and $\dim f^{-1}(y)=\dim X-\dim Y$ for all closed $y$ in a nonempty open subset of $Y$. If the requirement that $X$ and $Y$ are of finite type is dropped, and we just require them to be integral, separated, Noetherian schemes over an algebraically closed field, there are two ways I can see to interpret this claim:

(1) $\dim f^{-1}(y)\#\dim Y\geq\dim X$ for all closed $y\in Y$, and $\dim f^{-1}(y)\#\dim Y=\dim X$ for all closed $y$ in a nonempty open subset of $Y$, where $\dim$ refers to ordinal Krull dimension, and $\#$ is natural sum ($\alpha\#\beta$ is the largest ordinal that can be reached by interleaving $\alpha$ and $\beta$).

(2) $\text{codim}(f^{-1}(y)\text{ in }X)\geq\dim Y$ for all closed $y\in Y$, and $\text{codim}(f^{-1}(y)\text{ in }X)=\dim Y$ for all closed $y$ in a nonempty open subset of $Y$, where $\text{codim}(Z\text{ in }X)$ refers to the supremum of order types of chains of irreducible closed subsets of $X$ containing $Z$.

These two claims are not clearly equivalent in the infinite-dimensional setting. Are either or both of them true?

Edit: Friedrich Knop has shown that (1) is false, but his counterexample convinced me that I didn't ask the question I intended to. I am still interested in knowing whether this is true when $f$ sends constructible sets to constructible sets, whether it is true when there is an integral, separated, Noetherian scheme $Z$ such that $X\subseteq Z^{n+m}$ is constructible and $f$ is the projection map $X\rightarrow Y\subseteq Z^n$, and whether this example implies the condition that $f$ sends constructible sets to constructible sets.

## 1 Answer

(1) is wrong in this generality since the fiber dimensions can be different on dense subsets. Let, e.g., $R:=\mathbb C[x]$ and $S$ the localization of $R$ in $E$ where $E$ is the multipliciatively closed subset of all polynomials which do not have a zero in $\mathbb Q$. Put $X=\text{Spec }S$ and $Y=\text{Spec }R$. Then $X$ equals $Y$ with all irrational closed points removed. So $\dim X=\dim Y=1$. The preimage of a closed point $y\in Y$ is either a point if $y\in\mathbb Q$ ($\Rightarrow\dim f^{-1}(y)=0$) or empty ($\Rightarrow\dim f^{-1}(y)=-1$).

If you don't like empty fibers then take for $S$ the localization $R[y]_F=\mathbb C[x,y]_F$ where $F$ is the set of polynomials $p(x,y)$ which are (i) monic as a polynomial in $y$ and (ii) the constant term $p(x,0)$ has no rational zero. Then the points in $\mathbb Q\times\{0\}$ survive, so $\dim S=1$. But $\dim f^{-1}(y)$ is $1$ or $0$ depending on $y\in\mathbb Q$ or not.

(2) has a much higher chance of being true since for $Y=\mathbb C[x]$ this is just Krull's Hauptidealsatz.

• Ah, I was trying to say something that was true in the case where the dimensions were finite. Thanks for pointing out that that was wrong. – Alex Mennen May 24 '17 at 17:48