Let \(S\) be a scheme, let \(X\) be a reduced scheme over \(S\), and let \(Y\) be a separated scheme over \(S\). Let \(f\) and \(g\) be two \(S\dash\)morphisms from \(X\) to \(Y\) which agree on an open dense subset of \(X\). Show that \(f = g\).
Give examples to show that this result fails if either
\(X\) is nonreduced, or
\(Y\) is nonseparated.
Hint: consider the map \(h: X \to \fiberprod{Y}{S}{Y}\) obtained from \(f\) and \(g\).
Solution
1. Let\[ h=(f,g):X\longrightarrow Y\times_SY. \]The scheme-theoretic equalizer of \(f\) and \(g\) is\[ E=X\times_{Y\times_SY}Y, \]where the second map is the diagonal\[ \Delta_{Y/S}:Y\longrightarrow Y\times_SY. \]
For any \(S\)-scheme \(T\) with a morphism \(u:T\to X\), the map \(u\) factors through \(E\) exactly when \[ h\circ u \] factors through the diagonal. By the defining property of the diagonal, this is equivalent to \[ f\circ u=g\circ u. \] Thus \(E\) represents the equalizer of \(f\) and \(g\).
2. Because \(Y\) is separated over \(S\), the morphism\[ E\longrightarrow X \]is a closed immersion.
Separatedness means that \[ \Delta_{Y/S}:Y\longrightarrow Y\times_SY \] is a closed immersion. Closed immersions are preserved by base change, and \(E\to X\) is its base change along \(h\).
3. If \(U\subseteq X\) is the dense open subset on which \(f=g\), then\[ U\subseteq |E|. \]Consequently\[ |E|=|X|. \]
The equality \(f|_U=g|_U\) says exactly that the inclusion \(U\hookrightarrow X\) factors through the equalizer \(E\), so \(U\subseteq|E|\).
By 2, \(|E|\) is closed in \(X\). Since it contains the dense subset \(U\), it contains its closure \(X\).
4. Since \(X\) is reduced, the closed immersion \(E\hookrightarrow X\) with \(|E|=|X|\) is an isomorphism.
Let \[ \mathcal I\subseteq\mathcal O_X \] be the ideal sheaf defining \(E\). The equality of underlying spaces means \[ V(\mathcal I)=X. \] Affine-locally this says every element of the corresponding ideal lies in every prime ideal, hence is nilpotent. Thus \(\mathcal I\) is a sheaf of nilpotent ideals.
Because \(X\) is reduced, its nilradical is zero, so \[ \mathcal I=0. \] Therefore \(E=X\) as schemes.
5. Hence\[ \boxed{f=g.} \]
By 4, every morphism to \(X\) factors through the equalizer \(E=X\). In particular the identity of \(X\) does, and the equalizer property in 1 gives \(f=g\).
6. The conclusion can fail when \(X\) is nonreduced even if \(Y\) is separated.
Let \[ R=k[x,\epsilon]/(\epsilon^2,x\epsilon), \qquad X=\Spec R, \qquad Y=\mathbb A^1_k=\Spec k[t]. \] Define two morphisms \[ f,g:X\longrightarrow Y \] by the ring maps \[ f^\sharp(t)=0, \qquad g^\sharp(t)=\epsilon. \] They are distinct because \[ \epsilon\ne0\text{ in }R. \]
Let \[ U=D(x)\subseteq X. \] Since \(x\) becomes invertible on \(U\) and \(x\epsilon=0\), one has \[ \epsilon|_U=0. \] Hence \[ f|_U=g|_U. \]
The nilradical of \(R\) is generated by \(\epsilon\), so \[ |X|\cong\Spec k[x]. \] Under this homeomorphism \(D(x)\) is the usual punctured affine line, which is dense. Thus \(f\) and \(g\) agree on a dense open but are not equal.
The target \(Y=\mathbb A^1_k\) is separated, so the failure is exactly the nonreducedness of \(X\).
7. The conclusion can fail when \(Y\) is nonseparated even if \(X\) is reduced.
Let \(Y\) be the affine line with doubled origin, obtained by gluing two copies \[ Y_1\cong\mathbb A^1_k, \qquad Y_2\cong\mathbb A^1_k \] along the common open \[ \mathbb G_m=D(t) \] by the identity. The morphism \[ Y\longrightarrow S=\Spec k \] is not separated.
Take \[ X=\mathbb A^1_k \] and let \[ f:X\xrightarrow{\sim}Y_1\hookrightarrow Y, \qquad g:X\xrightarrow{\sim}Y_2\hookrightarrow Y \] be the two chart inclusions. The scheme \(X\) is reduced.
On the dense open \[ D(t)=\mathbb G_m\subseteq X, \] the maps agree because this is precisely the open along which the two copies were glued. But at \(0\) they land at the two distinct origins of \(Y\), so \[ f\ne g. \]
8. Q.E.D.
Steps <1>1–<1>5 prove the theorem; 6 and 7 give the requested counterexamples.