I have a flat morphism of relative dimension $d$. Is it true that a base change of this morphism is a flat morphism of the same relative dimension? We can assume that all schemes are of finite type over some field and equidimensional.
Algebraic Geometry – Base Change of Flat Morphism and Relative Dimension
algebraic-geometry
Related Solutions
Consider a scheme $X$ of finite type over a field $k$. Define the intersection of two closed subschemes of $X$ to be their fibered product. The formation of the intersection commutes with base change. Therefore, if two irreducible components of $X$ have nonempty intersection over $k$, they will have nonempty intersection over its algebraic closure. Of course, they may no longer then be irreducible, but obviously two of their irreducible components will have nonempty intersection.
Yes.
Definition of “morphism of pure relative dimension” in Conrad’s Grothendieck Duality and Base Change
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 $x$.
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.
Best Answer
Let $\pi\colon X\to Y$ be a flat morphism of finite type $k$-schemes, which has relative dimension $d$. Let $f\colon Y'\to Y$ be another morphism of schemes over $k$. We have to show that the pullback $\pi'\colon X'=X\times_Y Y'\to Y'$ is flat of relative dimension $d$. We know that $\pi'$ is flat, so now we have to show that for an arbitrary point $y'\in Y$, the fiber $\pi'^{-1}(y')$ has dimension $d$ over $k(y')$. We have the following diagram of pullbacks:
(I'm sorry for its comical size.) By pullback pasting, the large rectangle is also a pullback diagram. However, the bottom composite equals $\mathrm{Spec}\,k(y')\to\mathrm{Spec}\,k(f(y'))\to Y$, and hence $\pi'^{-1}(y')$ is isomorphic to the base change $\pi^{-1}(f(y'))\times_{k(f(y'))} k(y')$. We know that $\pi^{-1}(f(y'))$ is of dimension $d$ over $k(f(y'))$ as $\pi$ is flat of relative dimension $d$, so $\pi'^{-1}(y')$ is also $d$-dimensional over $k(y')$, because the pure dimension of finite type schemes over a field is preserved under any field extension.