Call \((X, \OO_X)\in \Sch\) reduced iff \(\OO_X(U)\) has no nilpotents for every open \(U\), and for \(A\in \Ring\) define \(A^{\red}\da A/\sqrt{0}\) to be \(A\) modulo its ideal of nilpotents.
Show that \(X\) is reduced iff for every \(p\in X\), the local ring \(\OO_{X, p}\) has no nilpotents.
Let \(\OO_X^{\red}\) be the sheafification of \(U \mapsto \OO_X(U)^{\red}\). Show that \(X_{\red}\da (X, \OO_X^{\red})\) is a scheme, and that there is a morphism of schemes \(X_{\red}\xrightarrow{\red} X\) inducing a homeomorphism \(\abs{X_{\red}}\to \abs{X}\) on underlying topological spaces.
Let \(X \xrightarrow{f} Y\in \Sch\) with \(X\) reduced. Show that there is a unique morphism \(X \xrightarrow{g} Y_{\red}\) such that \(f\) is the composition
Hint
A section that is zero in every stalk is zero, by the sheaf axiom.
Cover by affines.
\(\sqrt{0_{R}} \leq \mfp\) for every \(\mfp\in \Spec R\).
\(R_{\red}\da R/\sqrt{0_{R}}\) is a quotient, and localization commutes with quotients.
Maps \(R\to S\) with \(S\) reduced factor through \(R_{\red}\).
Sheafification has the same stalks, and an isomorphism on stalks is an isomorphism of sheaves.
Pushforwards of reduced sheaves are reduced.
Solution
Part (a).
\(\implies\): suppose \(X\) is reduced and \(\OO_{X, p}\) has a nonzero nilpotent germ \(s_p\) with \(s_p^n = 0\). Represent \(s_p\) by \(s\in\OO_X(U)\) for an open \(U\ni p\); since \(s^n\) has zero germ at \(p\), shrinking \(U\) gives \(s^n = 0 \in \OO_X(U)\), while \(s\neq 0\) because its germ is nonzero. So \(s\) is a nonzero nilpotent in \(\OO_X(U)\), a contradiction.
\(\impliedby\): suppose every \(\OO_{X,p}\) is reduced, and let \(s\in \OO_X(U)\) with \(s^n=0\). For each \(p\in U\) the germ \(s_p\) satisfies \(s_p^n = 0\) in \(\OO_{X,p}\), so \(s_p=0\). A section whose germs all vanish is zero by the sheaf axiom, so \(s = 0\).
Part (b). Let \(P\) be the presheaf \(U\mapsto\OO_X(U)^\red\), so \(\OO_X^\red\) is its sheafification, and let \(\rho\colon\OO_X\to\OO_X^\red\) be the quotient maps followed by sheafification. Stalks commute with quotients by the nilradical, so \(P_p=\OO_{X,p}^\red=(\OO_X^\red)_p\).
Let \(U=\Spec A\) be an affine open of \(X\) and \(j\colon\Spec A^\red\to\Spec A\) the morphism induced by \(A\to A^\red\). Every prime contains \(\sqrt0\), so \(j\) is a homeomorphism; identify the two spaces. For an open \(V\subseteq U\), the map \(\OO_U(V)\to\OO_{\Spec A^\red}(V)\) lands in a reduced ring, so it factors through \(P(V)\); this is a map of presheaves \(\ro{P}{U}\to\OO_{\Spec A^\red}\). On the stalk at \(\mfp\) it is \((A_\mfp)^\red\to(A^\red)_{\mfp}\), an isomorphism because localization commutes with quotients and \(\sqrt{0_{A_\mfp}}=(\sqrt{0_A})_\mfp\). The induced map \(\ro{\OO_X^\red}{U}\to\OO_{\Spec A^\red}\) is therefore an isomorphism of sheaves. So \(X_\red\) is covered by open sets isomorphic to affine schemes, and it is a scheme.
Let \(\red\colon X_\red\to X\) be the identity on spaces with \(\red^\#\da\rho\). Over each affine \(U=\Spec A\) it is \(j\), a morphism of schemes, so \(\red\) is a morphism of schemes, and its underlying map is the identity, a homeomorphism.
Part (c). Let \(f\colon X\to Y\) with \(X\) reduced, and let \(g\) have the same underlying map as \(f\), which is possible since \(\red\) is the identity on spaces.
For an open \(V\subseteq Y\), \(f^\#_V\colon\OO_Y(V)\to\OO_X(f^{-1}V)\) lands in a reduced ring, so it kills the nilpotents and factors through \(\OO_Y(V)^\red\). These factorizations form a map of presheaves from \(U\mapsto\OO_Y(U)^\red\) to the sheaf \(f_*\OO_X\), which by the universal property of sheafification factors uniquely through \(\OO_Y^\red\); call the result \(g^\#\). Then \(\red\circ g=f\). On the stalk at \(x\in X\), \(g^\#_x\colon\OO_{Y,f(x)}^\red\to\OO_{X,x}\) is induced by the local map \(f^\#_x\), and the maximal ideal of \(\OO_{Y,f(x)}^\red\) is the image of that of \(\OO_{Y,f(x)}\), so \(g^\#_x\) is local and \(g\) is a morphism of schemes.
For uniqueness, a morphism \(g'\) with \(\red\circ g'=f\) has underlying map \(f\), and on stalks \(g'^\#_x\circ\rho_{f(x)}=f^\#_x\). Each \(\rho_{f(x)}\) is surjective, so the stalk maps of \(g'^\#\) agree with those of \(g^\#\), and \(g'=g\).