Hartshorne II.2

2.1. Let $\varphi:A\to A_f$ be the natural homomorphism given by $a\mapsto a/1.$ Observe that $\varphi$ induces a bijection between the ideals of $A$ not containing $f$ and the ideals of $A_f$ via the identification $\mathfrak{a}\mapsto\varphi(\mathfrak{a})A_f$ with inverse $\mathfrak{b}\mapsto\varphi^{-1}(\mathfrak{b}).$ This identification also identifies prime ideals with prime ideals and therefore defines a bijection $\psi:D(f)\to\text{Spec}(A_f)$ . Moreover, by our observations above, $\psi$ and $\psi^{-1}$ each send closed sets to closed sets. Hence $\psi$ is a homeomorphism.

It remains to show that we have a morphism $\psi^{\#}:\mathscr{O}_{\text{Spec}(A_f)}\to\psi_*\mathscr{O}_X|_{D(f)}.$ Observe that for $g/f^{n}\in A_f$ the set $D(g/f^{n})\subset\text{Spec}(A_f)$ is equal to $D(g/1)\subset\text{Spec}(A_f)$ and $(A_{f})_{g/f^n}\cong A_{fg}.$ Note $D(g/1)=D(f)$ if $g\in (f).$ The basic open sets of $D(f)$ are precisely the sets $D(g)\cap D(f)$ for $g\in A$ and $D(g)\cap D(f)=D(f)$ when $g\in (f).$ Thus $\psi$ identifies the basic open sets of $D(f)$ with the basic open sets of $\text{Spec}(A_f).$ Moreover, $\mathscr{O}_X|_{D(f)}(D(g))\cong A_{fg}\cong\mathscr{O}_{\text{Spec}(A_f)}(D(g/1))$ and these isomorphisms are uniquely determined from the natural isomorphism $\Gamma(D(f),\mathscr{O}_X|_{D(f)})\cong \Gamma(\text{Spec}(A_f),\mathscr{O}_{\text{Spec}(A_f)}.$ These isomorphisms define $\psi^{\#}(D(g)):\mathscr{O}_{\text{Spec}(A_f)}(D(g/1))\to \psi_*(\mathscr{O}_X|_{D(f)})(D(g/1)=\mathscr{O}_X|_{D(f)}(D(g))$ and thus they glue to define $\psi^{\#}(U)$ for all open sets. Since each is an isomorphism, $\psi^{\#}_{\mathfrak{a}}$ is clearly a local homomorphism of local rings for each $\mathfrak{a}\in\text{Spec}(A_f).$

2.2. Let $\{U_\alpha\}$ be an affine open cover for $X$ with $U_\alpha\cong\text{Spec}(A^\alpha).$ Observe that $U\cap U_\alpha$ is an open subset of $U_\alpha$ and hence $U\cap U_\alpha=\bigcup_{\beta} U_{\alpha,\beta}$ such that for each $\beta$ we have $U_{\alpha,\beta}\cong D(f_{\alpha,\beta})\subseteq\text{Spec}(A_{f_\alpha})$ with $f_{\alpha,\beta}\in A^\alpha.$ From proposition 2.2 we know $D(f_{\alpha,\beta})\cong\text{Spec}(A^\alpha_{f_{\alpha,\beta}}).$ Therefore $U=\bigcup_{\alpha,\beta}U_{\alpha,\beta}$ and $\{U_{\alpha,\beta}\}$ is an affine open cover of $U.$ Therefore $(U,\mathscr{O}_X|_U)$ is a scheme.

Lemma 1. Suppose that $X$ is a topological space, $\mathscr{P}$ is a presheaf on $X$ , $\mathcal{B}$ is a basis for the topology on $X$ , and $\mathscr{P}$ is a $\mathcal{B}$ -sheaf. Then $\mathscr{P}^{+}$ is isomorphic to the sheaf generated by $\mathscr{P}$ as a $\mathcal{B}$ -sheaf. See Eisenbud proposition I-12 and exercise I-13 for information on $\mathcal{B}$ -sheaves.

It suffices to observe that $f(B):\mathscr{P}(B)\to\mathscr{P}^{+}(B)$ is an isomorphism for each set $B\in\mathcal{B}.$ Clearly the map $f(B)$ is surjective for each $B\in\mathcal{B}$ by definition and it is given by mapping a section $s\in\mathscr{P}(B)$ to the function $B\to\bigsqcup_{P\in B}\mathscr{P}_P$ given by $P\mapsto s_P$ (see proposition 1.2).

It remains to show that $f(B)$ is injective for each $B.$ Suppose that $s,t\in\mathscr{P}(B)$ are such that $f(B)(s)=f(B)(t).$ Then $s_P=t_P$ for all $P\in B.$ Thus for each $P\in B$ there exists an open set $U_P\subset B$ containing $P$ such that $s|_{U_P}=t|_{U_P}.$ Since $\mathcal{B}$ is a basis, we may pic a basis set $B_P\subset U_P$ containing $P.$ Therefore for each $P\in B$ there exists a basic open set $B_P\in\mathcal{B}$ containing $P$ such that $s|_{B_P}=t|_{B_P}.$ Therefore $s=t$ since $\mathscr{P}$ is a $\mathcal{B}$ -sheaf.

Corollary 2. Suppose that $X$ is a topological space, $\mathscr{F}$ is a sheaf on $X,$ and $\mathscr{P}$ is a presheaf on $X$ that forms a $\mathcal{B}$ -sheaf on some basis $\mathcal{B}$ of $X.$ If $f:\mathscr{P}\to\mathscr{F}$ is a map of $\mathcal{B}$ -sheaves (i.e. a collection of maps $f(B)$ for each $B\in\mathcal{B},$ compatible with restrictions) such that $f(B)$ is an isomorphism for each $B\in\mathcal{B},$ then there exists a induced map $\mathscr{P}^{+}\to\mathscr{F}$ that is an isomorphism of sheaves.

Recall that it suffices to define a map of sheaves on a basis. In our case we have a basis $\mathcal{B}$ and a collection of compatible isomorphisms $f(B):\mathscr{P}^{+}(B)\to\mathscr{F}(B)$ for each $B\in\mathcal{B},$ and therefore define an isomorphism of sheaves $\mathscr{P}^{+}\to\mathscr{F}.$

2.3.a. Suppose that $U\subseteq X$ is open and $s\in\mathcal{O}_X(U).$ If $s^n=0$ then $s_P^n=0$ for each $P\in U.$ Conversely, if $s_P^n=0$ for some $P\in U$ then there exists an open neighborhood $V\subseteq U$ such that $(s|_V)^n=0.$ Therefore a nilpotent section over an open subset exists if and only if a nilpotent germ exists at a stalk.

2.3.b. Let us first show that for any ring $A$ we have $(\text{Spec}(A))_{red}\cong \text{Spec}(A_{\text{red}}).$ Let $\mathcal{R}$ be the nilradical of $A$ and recall that $\mathcal{R}$ is the intersection of all the prime ideals of $A.$ Thus $\mathcal{R}$ is the unique prime ideal contained in all other prime ideals and if $\mathcal{R}\ne (0)$ then $(0)$ Is not prime. So natural map $\varphi:A\to A/\mathcal{R}=A_{\text{red}}$ gives a bijection between the prime ideals of $A.$ By proposition 2.3c we have a map $f:\text{Spec}(A/\mathcal{R})\to\text{Spec}(A)$ given by $f(\mathfrak{p})=\varphi^{-1}(\mathfrak{p})$ and $f^{\#}:\mathcal{O}_{\text{Spec}(A)}\to f_*\mathcal{O}_{\text{Spec}(A/\mathcal{R})}$ given by localizing $\varphi.$ We also know that $f(V(\mathfrak{a}))=V(\varphi^{-1}(\mathfrak{a})).$ Thus $f$ is a homeomorphism by our observations about the relationship between the ideals of $A$ and $A/\mathcal{R}.$ For each $D(g)\subset \text{Spec}(A)$ we have $f^{\#}(D(g)):A_g\to (A/\mathcal{R})_{\varphi(g)}$ and $(A/\mathcal{R})_{\varphi(g)}=(A_g)_{\text{red}}.$ Thus $f_*\mathcal{O}_{\text{Spec}(A/\mathcal{R})}(D(f))\cong \mathcal{O}_{\text{Spec}(A)}(D(f))_{\text{red}}.$

Now let $(\mathcal{O}_{\text{Spec}(A)})_{\text{red}}$ be the presheaf given by $U\mapsto \mathcal{O}_{\text{Spec}(A)}(U)_{\text{red}}.$ The restriction maps $(\mathcal{O}_{\text{Spec}(A)})_{\text{red}}(D(f)) \to(\mathcal{O}_{\text{Spec}(A)})_{\text{red}}(D(g))$ , that is $(A_{f})_{\text{red}}\to (A_{g})_{\text{red}},$ for $\sqrt{(g)}\subset\sqrt{(f)}$ is the map induced by the map $A_f\to A_g.$ Thus $(\mathcal{O}_{\text{Spec}(A)})_{\text{red}}$ forms a $\mathcal{B}$ -sheaf on the basis $\mathcal{B}$ of basic open sets, and in fact it is the same $\mathcal{B}$ -sheaf given by $f_*\mathcal{O}_{\text{Spec}(A/\mathcal{R})}$ above. Thus by corollary $2$ above we have that $(\mathcal{O}_{\text{Spec}(A)})_{\text{red}}^{+}$ is isomorphic to $f_*\mathcal{O}_{\text{Spec}(A/\mathcal{R})}.$

Suppose now that $X$ is an arbitrary scheme and $\{U_{\alpha}\}$ is an affine open cover of $X$ with $U_{\alpha}\cong\text{Spec}(A_\alpha).$ By our work above we see that $(U_{\alpha})_{\text{red}}\cong\text{Spec}((A_\alpha)_{\text{red}})$ for each $\alpha.$ The sheaves $(\mathcal{O}_X|_{U_\alpha})_{\text{red}}$ agree on basic affine open sets and thus on intersections. Therefore they glue to a sheaf $\mathcal{F}$ on $X$ . Moreover, for each $\alpha$ we see that $\mathcal{F}|_{U_\alpha}$ is clearly equal to the sheaf $((\mathcal{O}_X)_{\text{red}})|_{U_\alpha}$ given in the problem statement. Thus $\mathcal{F}\cong(\mathcal{O}_X)_{\text{red}}.$ The maps of affine schemes $(U_\alpha,\mathcal{O}_X)\cong\text{Spec}(A_\alpha)\to\text{Spec}((A_\alpha)_{\text{red}})\cong (U_\alpha,(\mathcal{O}_{X})_{\text{red}})$ also agree on intersections and hence glue to a map $X\to X_{\text{red}}.$ The map $X\to X_{\text{red}}$ is a homeomorphism as it is a homeomorphism on each $U_\alpha.$

2.3.c. We define the morphism $g:X\to Y_{\text{red}}$ by $g(\mathfrak{p})=h^{-1}(f(\mathfrak{p}))$ , where $h:Y_{\text{red}}\to Y$ is the morphism from part (b). Note that $h$ is a homeomorphism by part (b), so $g$ makes sense as function and is a continuous map of topological spaces. $\xymatrix{ X\ar[r]^{f}\ar[dr]_{g} & Y\\ & Y_{\text{red}} \ar[u]^{h} }$ To simplify notation let us identify the topological spaces $|Y|$ and $|Y_{\text{red}}|$ so the morphism $h:Y\to Y_{\text{red}}$ gives a map $h^{\#}:\mathcal{O}_{Y}\to(\mathscr{O}_Y)_{\text{red}}.$ We also have the map $f^{\#}:\mathcal{O}_Y\to f_*(\mathcal{O}_X)$ given by the morphism $f.$ Thus to complete our construction of a morphism $g$ with the desired properties it suffices to show that there exists a pullback map $g^{\#}:\mathscr{O}_{Y_{\text{red}}}\to g^*\mathscr{O}_X$ such that $f^{\#}=g^{\#}\circ h^{\#}.$

Observe that for $U\subseteq X$ open we have pullback morphism $f^{\#}(U):\mathscr{O}_Y(U)\to f_*\mathscr{O}_X(U)=\mathscr{O}_X(f^{-1}(U))$ . $\xymatrix{ \mathscr{O}_{Y}(U) \ar[d]_{h^{\#}(U)}\ar[dr]^{f^{\#}(U)} & \\ (\mathscr{O}_{Y})_{\text{red}}(U) \ar[r]_{g^{\#}(U)} & f_*\mathscr{O}_X(U). }$ Moreover, since $\mathscr{O}_X$ is reduced, $\mathscr{O}_X(f^{-1}(U))$ is reduced. So the kernel of $f^{\#}(U)$ must contain the nilradical of $\mathscr{O}_Y(U)$ . Thus it contains the kernel of $h^{\#}(U): \mathscr{O}_Y(U)\to\mathscr{O}_{Y_{\text{red}}}(U)$ . So there exists a unique homomorphism $g^{\#}(U):\mathscr{O}_{Y_{\text{red}}}(U)\to f_*\mathscr{O}_X$ such that $f^{\#}(U)=g^{\#}(U)\circ h^{\#}(U).$ The diagram also commutes with restriction $U\to V$ for $V\subseteq U$ and therefore defines a morphism $g^{\#}:\mathscr{O}_Y\to f_*\mathscr{O}_X$ such that $f^{\#}=g^{\#}\circ h^{\#}.$ With our identification of $|Y|$ and $|Y_{\text{red}}|$ we have that $f_*\mathscr{O}_X=g_*\mathscr{O}_X.$ Thus our construction above defines a morphism $g^{\#}:(\mathscr{O}_{Y})_{\text{red}}\to g_*\mathscr{O}_X$ that gives a local homomorphism of local rings and thus $(g,g^{\#})$ define a morphism of schemes.

2.4. Let us first show the result for affine schemes (this is actually already done in Hartshorne in proposition 2.3c).

Suppose that $\varphi:A\to B$ is a ring homomorphism. Then we may define a map $f:\text{Spec}(B)\to\text{Spec}(A)$ by $\mathfrak{p}\mapsto\varphi^{-1}(\mathfrak{p}).$ Observe that $f^{-1}(V(\mathfrak{a}))=V(\varphi(\mathfrak{a}))$ so $f$ is continuous. Observe that for $D(g)\subset\text{Spec}(A)$ we have $\mathscr{O}_{\text{Spec}(A)}(D(g))\cong A_f.$ Moreover, $\begin{aligned}[t] f_*\mathscr{O}_{\text{Spec}(B)}(D(g)) &=\mathscr{O}_{\text{Spec}(B)}(f^{-1}(D(g))\\ &=\mathscr{O}_{\text{Spec}(B)}(D(\varphi(g)))\\ &\cong B_{\varphi(g)}. \end{aligned}$ We therefore define $f^{\#}(D_f):A_f\to B_{\varphi(g)}$ to be the map induced by $\varphi:A\to B.$ These glue to define a morphism $f^{\#}:\mathscr{O}_{\text{Spec}(A)}\to f_*\mathscr{O}_{\text{Spec}(B)}$ with $f^{\#}(\text{Spec}(A))=\varphi.$

In order to show that $\varphi$ defines a morphism of schemes it remains to show that $f^{\#}_{f(\mathfrak{p})}:\mathscr{O}_{\text{Spec}(A),f(\mathfrak{p})}\to \mathscr{O}_{\text{Spec}(B),\mathfrak{p}}$ is a local homomorphism of local rings for each prime ideal $\mathfrak{p}\subseteq B.$ Observe that $\mathscr{O}_{\text{Spec}(A),f(\mathfrak{p})}\cong A_{\varphi^{-1}(\mathfrak{p})},$ $\mathscr{O}_{\text{Spec}(B),\mathfrak{p}}\cong B_{\mathfrak{p}},$ and $f^{\#}_{f(\mathfrak{p})}:A_{\varphi^{-1}(\mathfrak{p})}\to B_{\mathfrak{p}}$ is given by $s/t\mapsto\varphi(s)/\varphi(t).$ Thus $\begin{aligned}[t] (f^{\#}_{f(\mathfrak{p})})^{-1}(\mathfrak{m}_{\mathfrak{p}})&= \{s/t\in A_{\varphi^{-1}(\mathfrak{p})} \, : \, \varphi(s)/\varphi(t)\in\mathfrak{m}_{\mathfrak{p}}\}\\ &=\{s/t\in A_{\varphi^{-1}(\mathfrak{p})} \, : \, s\in\varphi^{-1}(\mathfrak{p})\}\\ &=\mathfrak{m}_{\varphi^{-1}(\mathfrak{p})}. \end{aligned}$ So $f^{\#}_{f(\mathfrak{p})}$ is a local homomorphism of local rings for each $\mathfrak{p}\in\text{Spec}(B)$ and $f^{\#}$ is a morphism of schemes.

Suppose conversely that $f:\text{Spec}(B)\to\text{Spec}(A)$ is a morphism of schemes. The first part of our work above shows that the homomorphism $f^{\#}(\text{Spec}(A)):A\to B$ determines the morphism $f^{\#}.$ Moreover, since $f^{\#}$ is a morphism of locally ringed spaces we must have $(f^{\#}_{f(\mathfrak{p})})^{-1}(\mathfrak{m}_{\mathfrak{p}}) =\mathfrak{m}_{f({\mathfrak{p}})}.$ But the map $f^{\#}_{f(\mathfrak{p})}:A_{f(\mathfrak{p})}\to B_{\mathfrak{p}}$ is just the localization of the global map $f^{\#}(\text{Spec}(A)):A\to B$ at $f(\mathfrak{p}).$ Therefore $f^{\#}(\text{Spec}(A))^{-1}(\mathfrak{p})=f(\mathfrak{p}).$ Thus the map $f$ is also determined by the homomorphism $f^{\#}(\text{Spec}(A)).$ Hence there is a bijection $\text{Hom}(\text{Spec}(B),\text{Spec}(A))\cong\text{Hom}(A,B).$

Now suppose that $X$ is an arbitrary scheme and $\varphi:A\to\Gamma(X,\mathscr{O}_X)$ is a homomorphism. Let $\{U_\alpha\}$ be an affine open cover of $X.$ The map $\varphi:A\to\Gamma(X,\mathscr{O}_X)$ composed with the restriction map $\text{res}_{X,U_\alpha}:\Gamma(X,\mathscr{O}_X)\to\Gamma(U_\alpha,\mathscr{O}_X)$ is a ring homomorphism $\text{res}_{X,U_\alpha}\circ\varphi:A\to\Gamma(U_\alpha, \mathscr{O}_X).$ Therefore $\text{res}_{X,U_\alpha}\circ\varphi$ determines a unique morphism of affine schemes $f_\alpha:U_\alpha\to\text{Spec}(A).$ Now let $\alpha,\beta$ be distinct indexes. We observe that $f_\alpha|_{U_\alpha\cap U_\beta}=f_\beta|_{U_\alpha\cap U_\beta}$ because for each affine open set $V\subset U_\alpha\cap U_\beta$ we have $(f_\alpha|_V)^{\#}(\text{Spec}(A))=\text{res}_{X,V}\circ\varphi$ , $(f_\beta|_V)^{\#}(\text{Spec}(A))=\text{res}_{X,V}\circ\varphi$ , and we know that since $V$ is affine these ring homomorphisms uniquely determine $f_\alpha|_V$ and $f_\beta|_V.$ Thus the $f_\alpha$ glue to a morphism of schemes $f:X\to\text{Spec}(A)$ corresponding to the homomorphism $\varphi.$

The argument above also shows that any morphism $f:X\to\text{Spec}(A)$ is determined by $f^{\#}(\text{Spec}(A)):A\to\Gamma(X,\mathscr{O}_X)$ since each morphism of affine schemes $f|_{U_\alpha}$ is determined by $\text{res}_{X,U_\alpha}\circ f^{\#}(\text{Spec}(A)).$ Thus there is a bijection $\text{Hom}(X,\text{Spec}(A))\cong\text{Hom}(A,\Gamma(X,\mathcal{O}_X)).$