Definition of “morphism of pure relative dimension” in Conrad's Grothendieck Duality and Base Change

Question

In p. 6 of the book from the title, we read

If $f:X\to Y$ is a locally finite type map of schemes, we define the relative dimension of $f$ at $x\in X$ to be $\dim\mathscr{O}_{X_{f(x)},x}$. If this function is identically equal to $n\geq 0$, we say that $f$ has pure relative dimension $n$ (i.e., non-empty fibers of $f$ are equidimensional with dimension $n$).

In other places I find the notion of a morphism of (pure) relative dimension $n$ defined as the one in between Conrad's parentheses (call this “definition $a$”); this seems to be the standard. However, this is not equivalent to “$\dim\mathscr{O}_{X_{f(x)},x}=n$ for all $x\in X$” (call this “definition $b$”). For instance, the morphism $\mathbb{A}_k^1\to\operatorname{Spec}k$ satisfies definition $a$ but not $b$.

My question is:

Can we fix Conrad's definition $b$ to make it equivalent to $a$?

For example, if $X$ and $Y$ are locally of finite type over a field $k$ and $f$ is a morphism over $\operatorname{Spec}k$, I think the condition “$\dim\mathscr{O}_{X_{f(x)},x}=n$ for all closed points $x\in X$” would be equivalent to definition $a$, or something like that. Nonetheless, I don't know if we can achieve something that works for an arbitrary morphism of schemes locally of finite type.

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The linked document by KReiser in the comments contains a correction by Conrad. There, the author argues two things: (i) that $\dim_xX_{f(x)}$ equals the maximal dimension of an irreducible component of $X_{f(x)}$ passing through $x\in X_{f(x)}$ and (ii) that “$\dim_xX_{f(x)}=n$ for all $x\in X$” is equivalent to definition $a$. My trouble now is that I don't know how to prove (i), and neither I know how to show (ii) even taking (i) for granted. Regarding how to show (i), KReiser has given hints in the comments, but I feel I'm still missing something. Regarding how to prove (ii) (taking (i) for granted), one only has to show that all nonempty fibers are equidimensional, but why are they?

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EDIT: I just found the necessary lemma in the SP, 0A21(5): for a scheme $X$ locally of finite type over a field, one has $\dim_x(X)=\max(\dim Z)$, where the maximum is over the irreducible components $Z$ of $X$ passing through $x$.

Answer 1

As mentioned in the comments and the edit, Brian Conrad's corrections to his book replace the objectionable definition with "the relative dimension of $f$ at $x$ ... should be the maximal dimension of an irreducible component of $X_{f(x)}$ passing through $x$..." (where $f:X\to Y$ is a morphism locally of finite type). The goal is to show that this is the same as $\dim_x X_{f(x)}$, the local dimension of $X_{f(x)}$ at $x$, defined as the minimum dimension of an open neighborhood of $X_{f(x)}$ containing $x$.

Now we have a scheme locally of finite type over a field. Let's calculate the dimension of any irreducible closed subscheme.

Lemma. If $X\to \operatorname{Spec} k$ is an irreducible $k$-scheme locally of finite type, then $\dim X=\dim U$ for every nonempty open $U\subset X$.

Proof sketch. We may immediately reduce to the case when $X$ is integral, as the reduction map is a homeomorphism. It suffices to prove this for affine open $U$ (ref). But for an integral affine $k$-scheme, the dimension is the transcendence degree of the field of rational functions, and that field is the stalk of the structure sheaf at the generic point, so the dimension of every nonempty affine open is the same. $\blacksquare$

Now we can show the equivalence of local dimension at a point $x\in X$ and dimension of largest irreducible component passing through $x\in X$ for $X$ a $k$-scheme locally of finite type: any open neighborhood of $x$ must contain an open subscheme of every irreducible component through $x$, and every such subscheme has the same dimension as the irreducible component by the lemma, showing $\dim_x X \geq \max_{Y\ni x, Y\text{ irred}} \dim Y$. Let's show that we can achieve equality: pick any affine open neighborhood $U$ of $x$, then using the fact that such an affine open neighborhood is noetherian and has finitely many irreducible components, we may pick an affine subneighborhood $U'$ which excludes all irreducible components not passing through $x$. Then the dimension of this neighborhood is equal to the maximum dimension of an irreducible component of $X$ passing through $x$, again using the lemma on the dimension of an open subset and the fact that irreducible components of $X$ passing through $x$ are in bijection with irreducible components of $U$ passing through $x$.

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To show the second part, let's recognize that we're trying to show that a scheme locally of finite type over a field has all irreducible components of dimension $n$ iff for all $x\in X$, the maximum dimension of an irreducible component through $x\in X$ is $n$.

Now this should be relatively straight forwards: from the lemma, we can check this affine-locally. If $X$ has an irreducible component not of dimension $n$, then there must be points on this irreducible component which are not on any other irreducible component of $X$ (the union of the finitely many other irreducible components is a closed set which must intersect this irreducible component in a proper subset). The local dimension at one of these points is therefore not $n$. The same approach run backwards proves the other direction.

Answer 2

I will just write in a more understandable way for myself and a more direct manner the proof of KReiser. (They prove first “$\geq$” and after “$\leq$” but actually the latter constitutes already a proof of “$=$”.) Along this answer, I will take for granted the following result (and I will not mention it when used):

If $X$ is a topological space and $U\subset X$ is open, then the assignments $T\subset X\mapsto U\cap T$ and $Z\subset U\mapsto \overline{Z}$ are the mutually inverse maps of a one-to-one correspondence between the closed irreducible subsets of $X$ meeting $U$ and the closed irreducible subsets of $U$. In particular, the assignments restrict to a one-to-one correspondence between the irreducible components of $X$ meeting $U$ and the irreducible components of $U$.

Lemma 1. Let $X$ be a topological space. Then $$ \dim X=\max\{\dim T\mid T\subset X\text{ irreducible component}\}. $$

(We are using this definition of dimension.)

Proof. ($\geq$). It is straightforward.

($\leq$). It follows from the fact that every irreducible subset of $X$ is contained in an irreducible component, 004W. $\square$

Lemma 2. Let $X$ be a scheme locally of finite type over a field.

1. $\dim T=\dim T\cap U$ for all irreducible components $T\subset X$ and all open subsets $U\subset X$ meeting $T$.

2. Let $x\in X$. Then $ \dim_xX=\max\{\dim T\mid T\subset X\text{ irreducible component, }x\in T\} $.

Proof. The proof of the first part follows from the lemma in KReiser's answer (in the SP it's 0A21(3)) plus the fact that all closed subsets of a scheme are the topological space of some closed subscheme, see 01J3.

We prove the second part. A scheme locally of finite type over a field locally has finitely many irreducible components, i.e., there is an open neighborhood $U\subset X$ of $x$ with finitely many irreducible components (for instance, $U$ could be any open affine neighborhood). There exists $V\subset U$ open neighborhood of $x$ such that all irreducible components of $V$ contain $x$ (take $V=$ the complement in $U$ of the union of the irreducible components of $U$ not containing $x$). Using part 1 and Lemma 1, we deduce $\dim W=\dim V$ for all open neighborhoods $W\subset V$ of $x$. Hence,

\begin{align*} \dim_xX &=\dim V\\ &=\max\{\dim Z\mid Z\subset V\text{ irreducible component}\}, &\text{by Lemma 1,}\\ &=\max\{\dim T\cap V\mid T\subset X\text{ irreducible component}\}\\ &=\max\{\dim T\cap V\mid T\subset X\text{ irreducible component, }x\in T\}\\ &=\max\{\dim T\mid T\subset X\text{ irreducible component, }x\in T\}, &\text{ by part 1. }\square \end{align*}