Lemma 26.12.4. Let $X$ be a scheme. Let $T \subset X$ be a closed subset. There exists a unique closed subscheme $Z \subset X$ with the following properties: (a) the underlying topological space of $Z$ is equal to $T$, and (b) $Z$ is reduced.

Proof. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the sub presheaf defined by the rule

$\mathcal{I}(U) = \{ f \in \mathcal{O}_ X(U) \mid f(t) = 0\text{ for all }t \in T\cap U\}$

Here we use $f(t)$ to indicate the image of $f$ in the residue field $\kappa (t)$ of $X$ at $t$. Because of the local nature of the condition it is clear that $\mathcal{I}$ is a sheaf of ideals. Moreover, let $U = \mathop{\mathrm{Spec}}(R)$ be an affine open. We may write $T \cap U = V(I)$ for a unique radical ideal $I \subset R$. Given a prime $\mathfrak p \in V(I)$ corresponding to $t \in T \cap U$ and an element $f \in R$ we have $f(t) = 0 \Leftrightarrow f \in \mathfrak p$. Hence $\mathcal{I}(U) = \cap _{\mathfrak p \in V(I)} \mathfrak p = I$ by Algebra, Lemma 10.16.2. Moreover, for any standard open $D(g) \subset \mathop{\mathrm{Spec}}(R) = U$ we have $\mathcal{I}(D(g)) = I_ g$ by the same reasoning. Thus $\widetilde I$ and $\mathcal{I}|_ U$ agree (as ideals) on a basis of opens and hence are equal. Therefore $\mathcal{I}$ is a quasi-coherent sheaf of ideals.

At this point we may define $Z$ as the closed subspace associated to the sheaf of ideals $\mathcal{I}$. For every affine open $U = \mathop{\mathrm{Spec}}(R)$ of $X$ we see that $Z \cap U = \mathop{\mathrm{Spec}}(R/I)$ where $I$ is a radical ideal and hence $Z$ is reduced (by Lemma 26.12.3 above). By construction the underlying closed subset of $Z$ is $T$. Hence we have found a closed subscheme with properties (a) and (b).

Let $Z' \subset X$ be a second closed subscheme with properties (a) and (b). For every affine open $U = \mathop{\mathrm{Spec}}(R)$ of $X$ we see that $Z' \cap U = \mathop{\mathrm{Spec}}(R/I')$ for some ideal $I' \subset R$. By Lemma 26.12.3 the ring $R/I'$ is reduced and hence $I'$ is radical. Since $V(I') = T \cap U = V(I)$ we deduced that $I = I'$ by Algebra, Lemma 10.16.2. Hence $Z'$ and $Z$ are defined by the same sheaf of ideals and hence are equal. $\square$

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