Let $X, Y, Z$ be schemes, where $X, Y$ are $Z$ schemes. I know the definition of "the locally closed subscheme of $X$ where two $Z$– morphisms $\pi, \pi': X\rightarrow Y$ agree" from its universal property. Also I can define it as the fiber product of the diagonal $$\delta : Y\rightarrow Y\times_Z Y$$ with $$(\pi, \pi'): X\rightarrow Y\times_Z Y.$$
My question: how to prove that the underlying set of "the locally closed subscheme where the two morphisms agree" is the same as the set of points where the two morphism agree on the residue field.
It is probably clear thatthe former is contained in the latter, but why is it all of them? That is, why is a point where $\pi, \pi'$ agree on the residue field necessarily contained in "the subscheme where $\pi, \pi'$ agree"?
Best Answer
Let "the set where the morphisms agree" be called $A$ and the fiber product $X\times_{Y\times_Z Y} Y$ be called $V$. Our goal is to show the underlying sets are equal. You state in your post that you are ok with $V\subset A$ set-theoretically (where we fudge things a little and identify $V$ with it's image under the immersion $V\to X$).
To show that $V=A$ set theoretically, it is enough to show that if $a\in A$, the fiber $V \times_X \operatorname{Spec} k(a)$ is nonempty. To start, we define the maps $i:\operatorname{Spec} k(a)\to X$ given by the standard inclusion and $j:\operatorname{Spec} k(a)\to Y$ given by $j((0))=f(a)$ as sets and $j^\sharp:\mathcal{O}_Y \to j_*\mathcal{O}_{\operatorname{Spec} k(a)}$ is given by $\mathcal{O}_Y(U)\to \mathcal{O}_{Y,f(a)} \to k(f(a)) \to k(a)$ if $a\in U$ and $0$ otherwise. As $i$ and $j$ agree when composed with the maps to $Y\times_ZY$, this gives a map $\operatorname{Spec} k(a)\to V$ so that the composite of this map with the natural projections $V\to X$ and $V\to Y$ agree with $i$ and $j$ by the universal property of the fiber product. But this means that there's a point of $V$ which maps to $a$ and we're done.