Compare properties of a ring homomorphism to the induced morphism on spectra.
Let \(A\) be a ring, \(X = \Spec A\), and \(f \in A\). Show that \(f\) is nilpotent if and only if \(D(f)\) is empty.
Let \(\phi: A \to B\) be a homomorphism of rings and let \(f: Y = \Spec B \to X = \Spec A\) be the induced morphism of affine schemes. Show that \(\phi\) is injective if and only if the map of sheaves \(f^{\sharp}: \OO_X \to f_* \OO_Y\) is injective. Show furthermore that in that case \(f\) is dominant, that is, \(f(Y)\) is dense in \(X\).
With the same notation, show that if \(\phi\) is surjective then \(f\) is a homeomorphism of \(Y\) onto a closed subset of \(X\), and \(f^{\sharp}: \OO_X \to f_* \OO_Y\) is surjective.
Prove the converse to (c): if \(f: Y \to X\) is a homeomorphism onto a closed subset and \(f^{\sharp}: \OO_X \to f_* \OO_Y\) is surjective, then \(\phi\) is surjective.
For part (d), consider \(X' = \Spec\qty{A/\ker \phi}\) and apply parts (b) and (c).
Solution
1. For \(f\in A\),\[ D(f)=\varnothing \iff f\text{ is nilpotent}. \]
By definition, \[ D(f)=\{\mathfrak p\in\Spec A:f\notin\mathfrak p\}. \] Thus \(D(f)=\varnothing\) exactly when \(f\) lies in every prime ideal of \(A\). The intersection of all prime ideals is the nilradical \(\sqrt{(0)}\), so this is equivalent to \(f\) being nilpotent.
2. Let \(\phi:A\to B\) be injective. Then\[ f^\sharp:\mathcal O_X\longrightarrow f_*\mathcal O_Y \]is injective.
The distinguished opens \(D(a)\) form a basis of \(X\). Their inverse images are \[ f^{-1}(D(a))=D(\phi(a))\subseteq\Spec B, \] and the map on sections is \[ A_a\longrightarrow B_{\phi(a)}. \] Suppose \(x/a^n\) maps to zero. Then for some \(m\ge0\), \[ \phi(a)^m\phi(x)=0. \] Injectivity of \(\phi\) gives \(a^mx=0\) in \(A\), hence \(x/a^n=0\) in \(A_a\). Thus the map is injective on every distinguished basic open, and therefore \(f^\sharp\) is injective as a map of sheaves.
3. Conversely, if \(f^\sharp\) is injective, then \(\phi\) is injective.
On global sections the sheaf map is exactly \[ \Gamma(X,\mathcal O_X)=A \longrightarrow \Gamma(Y,\mathcal O_Y)=B, \] namely \(\phi\). An injective sheaf morphism is injective on sections over every open set, in particular over \(X\). Hence \(\phi\) is injective.
4. If \(\phi\) is injective, then \(f:Y\to X\) is dominant.
It is enough to show that every nonempty distinguished open \(D(a)\subseteq X\) meets \(f(Y)\). By 1, nonemptiness of \(D(a)\) means that \(a\) is not nilpotent. Since \(\phi\) is injective, \(\phi(a)\) is not nilpotent in \(B\). Again by 1, \[ D(\phi(a))\ne\varnothing. \] Choose \(\mathfrak q\in D(\phi(a))\). Then \[ f(\mathfrak q)=\phi^{-1}(\mathfrak q)\in D(a). \] Thus the image meets every nonempty basic open, so it is dense.
5. If \(\phi\) is surjective and \(I=\ker\phi\), then\[ Y\cong\Spec(A/I) \]and \(f\) is a homeomorphism onto the closed subset \(V(I)\subseteq X\).
The first isomorphism theorem gives \[ B\cong A/I. \] Prime ideals of \(A/I\) correspond bijectively to prime ideals of \(A\) containing \(I\), and this correspondence is a homeomorphism \[ \Spec(A/I)\xrightarrow{\sim}V(I). \] Under the identification \(B\cong A/I\), this is exactly the map \(f\).
6. If \(\phi\) is surjective, then \(f^\sharp\) is surjective.
For a distinguished open \(D(a)\subseteq X\), the map on sections is \[ A_a\longrightarrow B_{\phi(a)}. \] Localization preserves surjections, so this map is surjective. Since distinguished opens form a basis, \(f^\sharp\) is surjective as a sheaf morphism.
7. Assume conversely that \(f\) is a homeomorphism onto a closed subset of \(X\) and that \(f^\sharp\) is surjective. Put\[ I=\ker\phi, \qquad X'=\Spec(A/I), \]and factor\[ Y\xrightarrow{g}X'\xrightarrow{h}X. \]Then \(g\) is a homeomorphism.
The ring map \[ A/I\longrightarrow B \] is injective. By 4, \(g(Y)\) is dense in \(X'\).
By 5, \(h\) is a homeomorphism of \(X'\) onto \(V(I)\). Also \[ f(Y)=h(g(Y)). \] The set \(f(Y)\) is closed by hypothesis, so \(g(Y)\) is closed in \(X'\). Since it is both closed and dense, \[ g(Y)=X'. \] Finally \(f\) and \(h\) are homeomorphisms onto the same closed subset \(V(I)=f(Y)\), and \(f=h\circ g\). Hence \[ g=h^{-1}\circ f \] is a homeomorphism.
8. The sheaf map\[ g^\sharp:\mathcal O_{X'}\longrightarrow g_*\mathcal O_Y \]is an isomorphism.
Because \(A/I\to B\) is injective, 2 shows that \(g^\sharp\) is injective.
For surjectivity, work on stalks. At a point \(y\in Y\), write \(x'=g(y)\) and \(x=h(x')\). The map on stalks for \(f=h\circ g\) factors as \[ \mathcal O_{X,x} \longrightarrow \mathcal O_{X',x'} \xrightarrow{g^\sharp_{x'}} \mathcal O_{Y,y}. \] The composite is surjective because \(f^\sharp\) is surjective. Therefore the second arrow is surjective as well. Thus \(g^\sharp\) is both injective and surjective on every stalk, hence is an isomorphism of sheaves.
9. Under the hypotheses of 7, \(\phi\) is surjective.
By 7, \(g\) is a homeomorphism, and by 8 its map on structure sheaves is an isomorphism. Hence \(g\) is an isomorphism of schemes.
Taking global sections gives an isomorphism \[ A/I\xrightarrow{\sim}B. \] Thus the original map \[ A\longrightarrow B \] is the quotient map by \(I=\ker\phi\), followed by an isomorphism, and is therefore surjective.
10. Q.E.D.
Step 1 proves part (a), steps <1>2–<1>4 prove part (b), steps <1>5–<1>6 prove part (c), and steps <1>7–<1>9 prove part (d).