Lemma 26.21.5. Let $X$, $Y$ be schemes over $S$. Let $a, b : X \to Y$ be morphisms of schemes over $S$. There exists a largest locally closed subscheme $Z \subset X$ such that $a|_ Z = b|_ Z$. In fact $Z$ is the equalizer of $(a, b)$. Moreover, if $Y$ is separated over $S$, then $Z$ is a closed subscheme.

Proof. The equalizer of $(a, b)$ is for categorical reasons the fibre product $Z$ in the following diagram

$\xymatrix{ Z = Y \times _{(Y \times _ S Y)} X \ar[r] \ar[d] & X \ar[d]^{(a , b)} \\ Y \ar[r]^-{\Delta _{Y/S}} & Y \times _ S Y }$

Thus the lemma follows from Lemmas 26.18.2, 26.21.2 and Definition 26.21.3. $\square$

Comment #7077 by Zeyn Sahilliogullari on

Is "locally closed subscheme" explicitly defined elsewhere in the stacks project?

Comment #7089 by Zeyn Sahilliogullari on

@#7077 "locally closed subscheme" is defined at the end of the section on scheme immersions:
https://stacks.math.columbia.edu/tag/01IM

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