Surjective closed morphism of schemes induces inequality of dimensions

Let $f: Y\to X$ be a surjective closed morphism of schemes. Prove $\dim Y\geq \dim X$.

I was made to believe that this should be easy, but I somehow did not manage to prove this. Using the fact that $f$ is closed and surjective, it is easy to reduce this to the case where both $X$ and $Y$ are irreducible. Now, if both were (locally) of finite type over a field, then the inequality would follow from the fact that the morphism induces an extension of the function fields. In the general case, however, I did not manage to produce a proof.

Clearly it would be enough to prove the following lemma:

Let $f:Y\to X$ be a closed morphism of schemes. Then for any $y\in Y$ and any generization $x_0$ of $x=f(y)$ there is a generization $y_0$ of $y$ with $f(y_0)=x_0$.

This lemma is true in the case where $f$ is an open morphism, but I don't know whether it is true in the closed case as well. At any rate, I also failed to prove this so far.

Any input on this problem would be highly appreciated, especially as I suspect that I am making things unneccessarily complicated right now.

Solution 1:

By the reduction claimed in your post, we may reduce the general case to the case where $X,Y$ are both irreducible. From here, we induct on $\dim X = n$. The case of $n=0$ is trivial. For the induction step, let $X_0\subset X_1\subset \cdots \subset X_n \subset X_{n+1} = X$ be a chain of proper inclusions of irreducible closed subsets of maximal length inside $X$. By definition of this chain, each $X_i$ is of dimension $i$. Then $f^{-1}(X_n)$ is a closed subset of $Y$, and as the restriction of closed and surjective maps is again closed and surjective, we may apply the induction hypothesis to see that $\dim f^{-1}(X_n)$ has dimension at least $n$. But $f^{-1}(X_n)$ is a proper closed subset of $Y$, and since $f^{-1}(X_n)$ is dimension at least $n$, it has an irreducible component of dimension at least $n$, which is properly contained in the irreducible set $Y$. Thus $\dim Y \geq n+1$, and we have proven the claim.

An alternate approach proceeds from the concept of specialization. In a topological space $X$, we say that $x$ is a specialization of $x'$ if $x\in\overline{\{x'\}}$ and denote this $x'\leadsto x$. The key fact for us is that for a scheme (more generally, any sober space) a specialization $x'\leadsto x$ is equivalent to an inclusion of irreducible closed subspaces $\overline{\{x'\}}\supset \overline{\{x\}}$ since every irreducible closed subset has a unique generic point. This means that we can calculate the dimension as the length of the longest chain of nontrivial specializations.

Now let's do something with specializations. Given a continuous map of topological spaces $f:Y\to X$, we say specializations lift along $f$ if given $x'\leadsto x$ in $X$ and any $y'\in Y$ with $f(y')=x'$, there exists a specialization $y'\leadsto y$ with $f(y)=x$. I claim that if $f$ is a closed map, then specializations lift along $f$: if $x'\leadsto x$ is a specialization and $f(y')=x'$, then $f(\overline{\{y'\}})$ is a closed subset containing $x'$, so it also contains $x$ and therefore there is some $y\in \overline{\{y'\}}$ with $f(y)=x$.

This now gives our result. For any chain of nontrivial specializations $x'\leadsto \cdots\leadsto x$ in $X$, we have that there is some $y'\mapsto x$ by surjectivity, and then since specializations lift along $f$, we get a corresponding chain of nontrivial specializations $y'\leadsto\cdots\leadsto y$ in $Y$, so $\dim Y\geq \dim X$.