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I am trying to learn about etale morphisms, and one definition that is given often is a smooth morphism of relative dimension $0$. I know that relative dimension can be defined generally in the case of flat morphisms, but there seems to be some confusion in the literature about the definition. Can anyone tell me definitively which of the following is actually meant by "relative dimension $n$":

Every non-empty fiber has dimension $n$, every non-empty fiber has pure dimension $n$, every fiber has dimension $n$, or every fiber has pure dimension $n$.

I have noticed some sources do say literally every fiber, but then a corollary of this is that any flat morphism is surjective, since otherwise some fibers would be empty. So it seems more likey that it is only referring to non-empty fibers.

Luke
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    The Stacks project uses (what I think is) the first definition: https://stacks.math.columbia.edu/tag/02NI – Qiaochu Yuan Dec 06 '20 at 23:36
  • Hi Qiaochu, thanks for the quick response. The stacks project was actually part of my confusion, since they use the term "equidimensional". I have seen that used in some places to mean "pure dimension", so it wasn't clear if they meant "equi" between fibers, or that each fiber itself is "equidimensional". – Luke Dec 06 '20 at 23:44
  • Unfortunately I don't know what the Stacks project means by "equidimensional" either... – Qiaochu Yuan Dec 06 '20 at 23:50
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    It's both - the goal statement is that every irreducible component of every fiber has the same dimension, $d$. Any fiber is equidimensional, in that every irred component of $X_y$ for fixed $y$ has the same dimension; then we ask that this dimension is the same for all $y$. – KReiser Dec 06 '20 at 23:54

2 Answers2

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At risk of inviting Cunningham's law down on myself, the definition I'm used to, have seen in the literature, and think is the "usual one" is the statement that a morphism of schemes $f:X\to Y$ is of relative dimension $d$ iff for all $y\in Y$, the fiber $X_y$ is equidimensional of dimension $d$. That is, all irreducible components of $X_y$ are of dimension $d$ for any $y$. In particular, we allow empty fibers since they have no irreducible components (an irreducible topological space is nonempty).

KReiser
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There's an equivalent definition to KReiser's one (see Lemma below). For completeness, I will include the proof of their equivalence. The proof arose from the discussion I maintained with KReiser here.

If $f:X\to S$ is a morphism of schemes and $s\in S$, we denote by $X_s$ to the scheme-theoretic fiber at $s$. Recall that given a topological space $T$, the Krull dimension of $T$ at $t\in T$ is defined as: $$ \dim_tT=\min\{\dim U\mid U\subset T\text{ open neighborhood of }t\}. $$

Lemma. Let $f:X\to S$ be a morphism of schemes locally of finite type, and let $d$ be a non-negative integer. Then $d=\dim_x X_{f(x)}$ for all $x\in X$ if and only if all non-empty fibers of $f$ are equidimensional of dimension $d$.

Exercise. A topological space $X$ is said to have locally finitely many irreducible components if each point of $X$ has a neighborhood with finitely many irreducible components. For such a space, show that if $Z\subset X$ is an irreducible component, then there is a non-empty open set $U$ of $X$ such that $Z$ is the unique irreducible component of $X$ that $U$ meets (hence, $U\subset Z$).

Proof. Let $x\in X$. The fiber $X_{f(x)}$ is locally of finite type over $\kappa(f(x))$.

$(\Leftarrow)$. The result follows from the fact that for any scheme $T$ locally of finite type over a field and $t\in T$, we have that $\dim_tT$ equals the maximum among the dimensions of irreducible components of $T$ passing through $t$ [A021].

$(\Rightarrow)$. Let $Z$ be an irreducible component of $X_{f(x)}$. Since $X_{f(x)}$ has locally finitely many irreducible components (it is locally Noetherian), by the exercise there is $x'\in Z$ such that $Z$ is the unique irreducible component of $X_{f(x)}$ passing through $x'$. By the fact invoked in last paragraph, $\dim Z=\dim_{x'} X_{f(x)}=\dim_{x'}X_{f(x')}=d$. $\square$

Elías Guisado Villalgordo
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