The <coincidence locus of two scheme morphisms> is the fibre product
$$
E=X\times_{(f,g),\,Y\times_kY,\,\Delta_Y}Y.
$$
The <diagonal morphism> $\Delta_Y$ is a <locally closed immersion>, and this property is stable under <base change of a morphism of schemes>, so $E\hookrightarrow X$ is locally closed. The <universal property of a fibre product> says that a morphism $T\to X$ factors through $E$ exactly when $f|_T=g|_T$. Thus $E$ is the largest locally closed subscheme on which they coincide. If $Y$ is <separated scheme>[separated], $\Delta_Y$ is a <closed immersion>, and its base change $E\hookrightarrow X$ is closed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 113 / 1 / d / Solution

For morphisms $f,g:X\to Y$ over a base $S$, their scheme-theoretic coincidence locus is the fibre product
$$
\operatorname{Eq}(f,g)=X\times_{(f,g),\,Y\times_SY,\,\Delta_{Y/S}}Y.
$$
It is the <equalizer> of $f$ and $g$, hence is universal among subschemes on which they agree. It is locally closed because every diagonal is an immersion, and closed when $Y\to S$ is separated.

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