Consider a scheme $(X,\mathcal{O}_X)$ and a closed subset $Y$ of $X$. Let $X = \bigcup U_i, \, U_i \cong \operatorname{Spec}(A_i)$ be an open affine covering of $X$ and let $Y_i = U_i \cap Y$. Then $Y_i$ is closed in $U_i$ and it can be endowed with a closed subscheme structure via the ring epimorphism $A_i \rightarrow A_i / a_i$, where $a_i = \bigcap_{P \in\operatorname{Spec}(A_i) \cap Y} P$. Notice that $a_i$ is radical and so $A_i/a_i$ is reduced. This construction endows each $Y_i$ with a sheaf $\mathcal{F}_i$ of reduced rings. Notice that the $Y_i$ form an open covering of $Y$. Thus we want to glue the $\mathcal{F}_i$ to get a reduced sheaf $\mathcal{F}$ on $Y$. For this, and in view of exercise II.$1.22$, we want to demonstrate isomorphisms $\phi_{ij}: \mathcal{F}_i|_{Y_i \cap Y_j} \rightarrow \mathcal{F}_j|_{Y_i \cap Y_j}$ as well as that the cocycle condition $\phi_{ik} = \phi_{jk} \circ \phi_{ij}$ holds on $Y_i \cap Y_j \cap Y_k$. Hartshorne says that this task can be reduced to the following: given affine open $U= \operatorname{Spec}(A)$ and $f \in A$, show that the reduced structure on $D(f) \cap Y$ induced by the restriction of the reduced structure on $\operatorname{Spec}(A) \cap Y$ is the same as the reduced structure on $\operatorname{Spec}(A_f) \cap Y$. Can somebody please explain what exactly is the argument that leads to this reduction?


1 Answer 1


$\require{AMScd}$ $\newcommand{\spec}[1]{\mathrm{Spec}(#1)}$ This reduction to the case $A \to A_f$ appears quite often in functorial constructions related to a scheme in algebraic geometry. It follows from the fact, that for a scheme $X$ and and a covering $(U_i)_i$ with open affines $U_i = \spec{A_i}$ we can choose a covering consisting of the original $U_i$ and additional affine $U_{ijk} = \spec{A_{ijk}} \in U_i \cap U_j$, such that the $(U_{ijk})_k$ cover $U_i \cap U_j$ and we have diagrams $$ \begin{CD} A_i @>>> A_i\\ @VVV @VVV\\ A_{ijk} @>>> (A_i)_{f_{ijk}} \end{CD} $$ and $$ \begin{CD} A_j @>>> A_j\\ @VVV @VVV\\ A_{ijk} @>>> (A_j)_{g_{ijk}} \end{CD} $$ where the vertical maps induce the respective open immersions and the horizontal ones are isomorphisms.

To prove this, we begin with a lemma:

Lemma A

Let $U=\spec{A}$ and $V = \spec{B}$ and $j:V \to U$ an open immersion induced by $\varphi:A \to B$. Furthermore be $D(f) = \spec{A_f} \subseteq j(V)$ be an open immersion too. Then $\spec{A_f} = \spec{B_{\varphi(f)}}$.

I do not prove this, it is fairly obvious.

Now for an $x \in U_i \cap U_j$ choose an $f'_{ijk}$ such that for $B = (A_i)_{f'_{ijk}}$ we have $x \in \spec{B} \subseteq U_i \cap U_j$. Then choose a $g'_{ijk}$, such that $x \in \spec{(A_j)_{g'_{ijk}}} \subseteq \spec{B}$. If $\psi:A_j \to B$ is the map inducing $\spec{B} \subseteq U_j$, then $\spec{B_{\psi(g'_{ijk})}} = \spec{(A_j)_{g'_{ijk}}}$ by the Lemma A. Now $B_{\psi(g'_{ijk})} = (A_i)_{f_{ijk}}$ for a certain $f_{ijk} \in A_i$. Concretely let $\psi(g'_{ijk}) = h_{ijk}/(f'_{ijk})^d$, then $f_{ijk} = f'_{ijk} h_{ijk}$ with $h_{ijk} \in A_i$. Setting $g_{ijk}=g'_{ijk}$ we have

$$x \in U_{ijk}:=\spec{A_{ijk}}:= \spec{(A_i)_{f_{ijk}}} = \spec{(A_j)_{g_{ijk}}} \subseteq U_i \cap U_j$$

So we have, depending on $x \in U_i \cap U_j$ constructed our $U_{ijk}$ that is a standard-open subset of $U_i$ as well as of $U_j$. Going over all $i,j$ and all $x$ we construct the totality of all the $U_{ijk}$.

In the covering $(U_i, U_{ijk})$ only the morphisms $\gamma_{i,ijk}:U_{ijk} \to U_i$ and $\gamma_{j,ijk}:U_{ijk} \to U_j$ exist and are by construction of the form $A \to A_f$. Gluing this data together gives the original scheme $X$.

Now for your specific case I would argue, that the functor

$$F(U) = (U \cap Y)_\mathrm{red}$$

respects isomorphisms $U \cong U'$ and (to be proved) we have $F(U_f) = F(U)_f$. So the glueing data from above gives gluing data $F(U_i), F(U_{ijk}), F(\gamma_{i,ijk}), F(\gamma_{j,ijk})$ which glue together to give $Y$ with the reduced induced structure.


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