Morphism between thick fibers of schemes extends to a neighbourhood

Question

Let $ S $ be a locally Noetherian scheme, and $ X $, $ Y $ finite type $ S $-schemes. Let us fix $ s \in S $. Let $ \varphi : X \times _ { S } \mathcal{O}_{S,s} \to Y \times _ { S } \mathcal{O}_{S,s} $ be a morphism of $ S $-schemes. Show that there exists an open subset $ W \ni s $ of $ S $ and a morphism $ f : X \times _ { S } W \to Y \times _ { S } W $ such that $ \varphi $ is obtained from $ f $ via base change $ \text{Spec} \mathcal{O}_{S,s} \to W $. If $ \varphi $ is an isomorphism, show that there exists such an $ f $ which is moreover an isomorphism. $ \quad $

P.S. This question is Exercise 2.3.5 from Qing Liu's book and is related to "Extending a morphism of schemes", "Extending a morphism from Spec $\mathcal{O}_{X,x}$". I am writing the solution below in order to record some of the details that initially trumped me.

Answer 1

Let $ p : X \to S $ and $ q : Y \to S $ be the structure maps. We can replace $ S $ with any open affine $ W $ around $ s $, $ X $ with $ p ^{-1} ( W ) $ and $ Y $ with $ q ^ { - 1 } ( W ) $ while the hypothesis still holds. Henceforth, assume that $ S $ is the spectrum of a Noetherian ring $ R $ and $ s $ is the prime $ \mathfrak{p} $ in $ R $. Set $ L = R \setminus \mathfrak{p} $.

Claim 1. The scheme $ X \times O _{ S,s }$ is topologically a subspace of $ X $ that is the union of all the fibers $ f^{-1}(t) = X_{t} = X \times _ { S } k(t) $ where $ t $ is a generization of $ s $ i.e. $ s \in \overline { \left \{ t \right \} } $. Moreover, any open subscheme of $ X \times O_{S,s} $ is isomorphic to $ U \times_{s} \mathcal{O}_{S,s} $ for some open subscheme $ U $ of $ X $.

Proof. Let us denote the underlying topological space of $ X \times _ { S } \mathcal{O}_{S,s} $ by $ T $.

For the first part, it suffices to show the claim when $ X $ is affine (Why?). Suppose $ X = \text{Spec } A $. Then, $ X \times _ { S } \mathcal{O}_{S,s} = \text{Spec } ( A \otimes R _ { \mathfrak{p} } ) = \text{Spec } ( L ^ { - 1 } A ) $. Since this is a localization, the space $ T $ is a subspace of $ X $. Moreover, $ T $ consists of exactly those primes $ \mathfrak{q} $ of $ A $ whose inverse image under the ring map $ R \to A $ is a prime $ \mathfrak{p} ' \subset \mathfrak{p} $ i.e. $ T $ is the pre-image of those points in $ S $ which are generizations of $ \mathfrak { p } $. This proves the first half of the claim.

Suppose now that $ V $ is an open subscheme of $ X \times _{S } \mathcal{O}_{S,s} $. Then, from the proof above of the first part, $ V $ as a topological susbspace is equal to $ U \cap T $ for some open subset $ U $ of $ X $. Now, $ U $ has a unique open sub-scheme structure inherited from $ X $ and the scheme $ U \times _ { S } \mathcal{O}_{S,s} $, which is an open subscheme of $ X \times _ { S } \mathcal{O}_{S,s} $, has the same underlying topological space as $ V $. Since the open subscheme structures are unique, we must have $ V \cong U \times _ { S} \mathcal{O}_{S,s} $. $ \square $

Henceforth, we shall call the scheme $ X \times _ { S } \mathcal{O} _ { S, s} $ the thick fiber of $ X $ at $ s $ and denote it by $ ^ t X _ { s } $. So, we are given a map $$ \varphi :\ ^{t}{X} _ { s} \to \ ^ t Y _ { s } . $$ and the problem is to "expand this map" around the subspaces $ ^t{X} $, $ ^{t}Y $ to $ p^{-1}(U) $, $ q^{-1}(U) $ for some open affine $ U \ni s $ in $ S $.

Since $ X $, $ Y $ are finite type $ R $-schemes, both $ X $ and $ Y $ are quasi-compact. Let $ U = \text{Spec } A $ be an aribrary open affine subset of $ X $ and $ V _{i} = \text{Spec } A _{ i } $, $ 1 \leq i \leq n $ be a finite open affine cover of $ Y $, and $ V_{i} ' = \text{Spec } A_{i} \otimes R_{\mathfrak{p}} $ be the corresponding covering of the thick fiber of $ Y $ at $ s $. We can cover $ U \cap \varphi ^ { - 1 } ( V_{i} ' ) $ for each $ i = 1, \ldots, n $ by finitely many distinguished open subsets of $ \text{Spec } L^{-1} A = U \times _ { S } \mathcal{O}_{S,s} $, say altogether by $$ U_{j} ' = \text{Spec } A_{\mathfrak{p}, g_{j}} \text{ for } 1 \leq j \leq N , \quad g_{j} \in A _ { \mathfrak{p} } $$ so that each of these open affines lands inside $ V_ { i } ' $ for some $ i $. Then, since $ D(g_{j}) $ cover $ \text{Spec } L ^ { -1 } A $, $ g_{j} $ generate the unit ideal in $ A _ { \mathfrak{p} } $. Suppose that $$ g_ { j } = \frac{x_{j } } { y_{ j } } $$ where $ x_{j} \in A $, $ y_{j} \in L $. Then, from the remark just made, there are $ \frac { a _{j} } { z_{j} } \in L ^ { - 1 } A $ where $ a_{j} \in A $, $ z_{j} \in L $ for $ j = 1 , \ldots, N $ such that $$ \sum_{i} \frac{ a _{i} } { z_{i} } \cdot \frac{x_{i} } { y_{i} } = 1 , $$ which, after clearing all denominators, implies that the ideal generated by $ x_{1} , \ldots , x_{N} $ in $ A $ is generated by the image of an element $ \alpha \in \eta ( L ) $, where $ \eta : R \to A $ is the ring map. If we set $ U_{j} = \text{Spec } A_{x_{j} } $, then we cannot say that $ U_{j} $ necessarily make a cover of $ U $. If however one replaces $$ R \rightsquigarrow R_{ \alpha } , $$ (and $ X $, $ Y $, $ V_{i} $, $ U $ by appropriate open subschemes) to begin with, $ U_{j} $ can now be assumed to cover $ U $ (since $ x_{i} $ now generate the unit ideal in $ A $).

Since $ X $ is quasi-compact, we can do the same argument for each open subset in a finite cover of $ X $. We thus end up with an affine cover $ \text{Spec } A_{j} $ of $ X $ for $ 1 \leq j \leq m $, a cover $ \text{Spec } B_{i} $ of $ Y $ for $ 1 \leq i \leq n $, and a finite collection of ring maps $ B_{i} \otimes R _ { \mathfrak { p } } \to A_{j} \otimes R _ { \mathfrak{p} } $ for pairs $ (i,j) $ in some set $ I \subset \left {1, \ldots, n \right \} \times \left\{ 1, m \right \} $ that agree on intersections.

Claim 2. Let $ A $ and $ B $ be $ R $-algebras where $ R $ is a Noetherian ring and let $ \mathfrak{p} $ a prime of $ R $. Suppose that we have a ring map $ \phi : B \otimes R_{ \mathfrak{ p } } \to A \otimes R _ { \mathfrak{p} } $. If $ B $ is finitely generated, there exists a $ r \in R \setminus \mathfrak{p} $ such that $ \phi $ is obtained from a map $ \psi : B _ { r } \to A_{ r } $ by localizing to $ R \setminus \mathfrak{p} $. Moreover, if $ r' $ and $ \psi ' : B _ { r '} \to A_{r' } $ are any other such pair, then the maps so defined from $ B_{rr' } \to A_{rr' } $ by localizing $ \psi , \psi ' $ are the same.

Proof. Suppose that $ B = R[t_{1}, \ldots, t_{n} ] / ( b_{1} , \ldots, b _ { m } ) $ where $ b_{1}, \ldots, b_{m} \in R [t_{1}, \ldots, t_{n} ] $. Consider the induced map $ \chi : R _ { \mathfrak{p} } [ t_{1} , \ldots, t_{n} ] \to A_{ \mathfrak{p} } $. Suppose that $$ t_{i} \mapsto \frac{a_{i}}{s_{i}} \quad \quad \quad \text{for } i = 1, \ldots, n $$ and let $ s = s_{1} s_{2} \ldots s_{n} $. Define a map $ \chi ' : R_{s} [ t_{1} , \ldots, t_{n } ] \to A_{ s } $ by $$ t_{i} \mapsto \frac{a_{i} s_{1} \cdots s_{i-1} s_{i+1} \cdots s_{n} } { s } \quad \quad \quad \text{ for } i = 1 , \ldots, n $$ and suppose that under $ \chi ' $, $$ b_{j} \mapsto \frac{ a_{j} ' } { s_{j} ' } \text{ for } j = 1, \ldots, m . $$ It is clear that $ \chi $ is obtained from $ \chi ' $ after localizing to $ R \setminus \mathfrak{p} $. Thus, each $ \frac{ a_{j} ' } { s _{j} ' } $ becomes a zero in $ A _ { \mathfrak { p } } $, which means that there are $ r_{j} \in R \setminus \mathfrak{p} $ such that $ r _{j} \cdot a_{j} ' = 0 $. Thus, taking $ r = s \cdot r _{1} r _{2} \cdots r _ {m} $, we see that the induced map $$ R_{r} [ t_{1} , \ldots , t_{n} ] \to A_{r} $$ sends $ b_{j} $ to zero, and thus further induces a map $$ \psi : B _ { r } \to A_{ r } . $$ This map $ \psi $ clearly localizes to $ \phi $ and is the desired map. The fact that any two such maps are compatible is straightforward. $ \square $

In light of Claim 2 above, we can find finitely many $ r_{ij} $ for $ (i, j ) \in I $ such that the maps $ B_{j} \otimes R_{ \mathfrak{p} } \to A_{i} \otimes R _ { \mathfrak { p} } $ are localizations of some maps $$ B _ { j } \otimes R _ { r _ { ij } } \to A_{i} \otimes R _ { r _ { i j } } $$ which agree on common open subsets. By taking the product of $ r_ { i j } $ to be $ r $, we obtain a common neighbourhood $ W = D(r) $ in $ \text{Spec } R $ and compatible maps $ B_ { j } \otimes R_{r} \to A_{i} \otimes R_{r} $. This amounts to giving a map $$ p^{-1} ( W ) \to q ^ {- 1} ( W ) $$ and which is the desired claim.

Remark 1. We only needed the map $ X \to S $ to be quasi-compact instead of finite type in the proof.

It is interesting to note that no hypothesis on $ X $ was needed in Extending a morphism of schemes, but is crucial here for the result to hold. The necessity of some hypothesis on $ X $ was pointed out to me Arnav Tripathy and Koji Shizimu and an explicit counterexample given by the latter is reproduced below.

Take $ S = \text{Spec } \mathbb{Z} $, $ s $ the generic point of $ S $, $ Y = \text{Spec } \mathbb{Z} [t] $ and $ X = \bigsqcup _ { i \in \mathbb{N} } X_{i} $ where $ X_{i} = \text{Spec } \mathbb{Z} [ t] $. Then, $ X \times \mathcal{ O } _{S,s} = \text{Spec } \mathbb{Q} [t ] $ and $ X \times _ { S } \mathcal{O} _{ S, s } = \bigsqcup_{ i \in \mathbb{N} } \mathbb{Q} [ t ] $. We can then define a map $ \varphi : X \times _ { S } \mathcal{O}_{S,s} \to Y \times \mathcal{O}_{S,s} $ by defining the map on the $ i $-th copy of $ \mathbb{Q} [ t] $ to $ \mathbb{Q} [ t] $ which sends $ t $ to $ t / i $. Since the open subsets of $ \text{Spec } \mathbb{Z} $ are of the form $ \text{Spec } \mathbb{Z} [ 1 / N ] $, it is impossible to find an open subset $ U $ of $ S $ such that $ \varphi $ is obtained from some $ X \times U \to Y \times U $, as there are infinitely many denominators involved.

Remark 2. The claim that open subsets of $ X \times \mathcal{O}_{S,s} $ are of the form $ U \times \mathcal{O}_{S,s} $ isn't true if $ \mathcal{O} _ { S , s } $ is replaced by an arbitrary $ S $-scheme $ Z $. See the counterexample here.

Remark 3. One may wonder that if any open subset of $ X \times \mathcal{O}_{S,s} $ is of the form $ U \times _{ S} \mathcal{O}_{S,s} $ for some open subset $ U $ of $ X $, how does one write $ X \times_{S} W $ in the said form where $ W $ is an open subset of $ \mathcal{O}_{S,s} $? This may seem counter-intuitive as first, but we can proceed as follows.

Since $ \text{Spec } \mathcal{O}_{S,s} $ is a subspace of $ S $ (it is, in fact, the intersection of all open subsets of $ S $ containing $ s $), any open subset $ W $ of $ \text{Spec } \mathcal{O}_{S,s} $ is obtained by intersecting some open subset $ V $ of $ S $ with this subspace. Let $ U = p ^ { - 1 } ( V ) $. Then, $ X \times _ { S } W = U \times _ { S } \mathcal{O}_{S,s} $.

Answer 2

Slightly less pedantic version of the same solution for quicker future reference.

We can assume that $ X $ and $ S $ are affine, say $ X = \text{Spec } A $ and $ S = \text{Spec } R $, since $ X $ is quasi-compact and we can shrink $ S $ to an open subset any finite number of times. Denote $ s $ by the prime $ \mathfrak{p} \subset R $. Let $ L = R \setminus \mathfrak{p} $. If $ Y $ is affine as well, say $ Y = \text{Spec } B $, the problem just boils down to showing that a map $ L ^ { - 1 } B = B \otimes R_ { \mathfrak{p} } \to A \otimes R _ { \mathfrak { p } } = L ^ { - 1 } A $ actually comes from (i.e. is the localization of) a map $ B_{r} \to A_{r} $ for some $ r \in L $ which is not hard using finiteness conditions on $ B $.

In general, we can choose a finite open affine cover $ V_{i} = \text{Spec } B_{i} $ of $ Y $ for $ i = 1, \ldots, n $. The inverse images $ U_{i} ' = \varphi^{-1} ( \text{Spec } ( L ^ { -1 } B_{i} ) ) $ can be covered with a finite number of open affines of $ \text {Spec } L ^ {- 1 } A $, which in turn, can be covered with principal open affines. So, altogether we can choose a collection of $ g_{j} \in L ^ { - 1 } A $ for $ j = 1 , \ldots, m $ such that $ \text{Spec } ( L ^ { - 1 } A ) _ { g_{j } } $ cover $ \text{ Spec } L ^{-1} A $ and each such open set lands in some $ \text{Spec } { L ^ { - 1 } B_{i} } $ under $ \varphi $. Suppose $ g_{j} = \frac{x_{j} } { y_{j} } $ for $ x_{j} \in A $, $ y_{j} \in L $. As $ y_{j} $ is invertible in $ L ^{-1} A $, we can assume $ y_{j} = 1 $. From the commutative algebra trick above, the maps $$ \text{Spec } L ^{-1} ( A _ { x_{j} } ) \to \text{Spec } L ^ { - 1 } B_{i} $$ actually come from some maps $$ ( B_{i} ) _ { r_{ij} } \to ( A_{x_{j} } ) _ {r_{ij} } \quad r_{ij} \in T $$ by localizing at $ T $. By taking the product $ r $ of all the $ r_{ij} $, we see that we have maps $$ \left ( \bigcup _ { j } \text{ Spec } A_{x_{j} } \right ) \times_{S} \text{ Spec } R_ { r } \to \left ( \bigcup _ { i } \text {Spec } B _ { i } \right ) \times _ {S} \text{ Spec } R_{r} $$ Now, the union of $ \text{ Spec } B_{j} $ is $ Y $ by definition, but $ \bigcup _ { i } \text{Spec } A _ { x_{i} } $ is not necessarily a cover of $ X = \text{Spec } A $. However, we can fix this as follows.

Inside $ A $, we have $$ D( x_{1} ) \cup D(x_{2} ) \cup \ldots \cup D(x_{n} ) = D( x_{1}, x_{2}, \ldots, x_{n} ) = D( \alpha ) $$ for some $ \alpha \in R $, since $ x_{i}/1 $ generate the unit ideal in $ L ^ { - 1 } A $. By further localizing $ R $ at $ \alpha $, we can make sure that $ D(x_{i} ) $ forms a cover of $ \text{Spec } A $.

Answer 3

$\DeclareMathOperator{\Spec}{Spec}$We may shrink $S$ to any open neighborhood of $s$. Let's first deal with the case everything is affine.

Claim 1: Let $R$ be a Noetherian ring and $\mathfrak{p}\in\Spec R$, $A,B$ be two finite type $R$-algebras, and $\varphi:A_{\mathfrak{p}}\to B_{\mathfrak{p}}$ be a map of $R_{\mathfrak{p}}$-algebra. Then there exists $f\in R\backslash\mathfrak{p}$ and an $R_{f}$-algebra map $\varphi^{\prime}:A_{f}\to B_{f}$ s.t. $\varphi^{\prime}\otimes\mathrm{id}_{R_{\mathfrak{p}}}=\varphi$. If $(g,\psi)$ is another such pair, then there exists $h\in R\backslash\mathfrak{p}$ s.t. the maps $A_{fgh}\to B_{fgh}$ induced by $\varphi^{\prime}$ and $\psi$ agree. Moreover if $\varphi$ is an isomorphism, then we can pick $(f,\varphi^{\prime})$ s.t. $\varphi^{\prime}$ is an isomorphism.

Proof: Since $R$ is Noetherian and $A$ is of finite type, $A$ must be finitely presented. Say $A=\frac{R[T_{1},\dots,T_{n}]}{(g_{1},\dots,g_{m})}$. Then $\varphi$ is uniquely determined by a finite set of relations: $\varphi(T_{i})=\frac{b_{i}}{r_{i}}\in B_{\mathfrak{p}}$ and $\varphi(g_{i})=0\in B_{\mathfrak{p}}$. We can lift those relations to $B_{f}$ for some $f\in R\backslash\mathfrak{p}$ and hence lift $\varphi$ to $\varphi^{\prime}:A_{f}\to B_{f}$. Similarly if $(g,\psi)$ is another such pair, we have $\forall i,\varphi^{\prime}(T_{i})=\psi(T_{i})$ in $B_{\mathfrak{p}}$, and we can lift the relations to $B_{fgh}$ for some $h\in R\backslash\mathfrak{p}$. It follows that $\varphi^{\prime}=\psi$ over $R_{fgh}$.

If $\varphi$ is an isomorphism with inverse $\psi=\varphi^{-1}:B_{\mathfrak{p}}\to A_{\mathfrak{p}}$. Then there exists $(f,\varphi^{\prime})$ for $\varphi$ and $(g,\psi^{\prime})$ for $\psi$. By above proof we may assume $f=g$. Now $\psi^{\prime}\circ\varphi^{\prime}=\mathrm{id}_{A_{f}}$ and $\varphi^{\prime}\circ\psi^{\prime}=\mathrm{id}_{B_{f}}$ over $R_{\mathfrak{p}}$, so there exists $g\in R\backslash\mathfrak{p}$ s.t. the equalities hold over $R_{fg}$. The result follows.$\square$

We will also use Claim 2 and Claim 3 which are stated and proved at the end of this answer.

Notation: for any $S$-scheme $Z$, denote $Z\times_{S}\Spec\mathcal{O}_{S,s}$ by $Z^{\prime}$. For any integer $m\geq1$, denote $\{1,\dots,m\}$ by $[m]$. If $T_{i}$ and $T_{j}$ are open in a scheme $T$, denote $T_{ij}:=T_{i}\cap T_{j}$.

Next we want to show the following: after shrinking $S$ finitely many times,

1. there exists finite open affine coverings $\{X_{j}\}_{j\in[m]},\{Y_{i}\}_{i\in[n]}$ for $X$ and $Y$ respectively and a function $\sigma:[m]\to[n]$ s.t. $\forall j\in[m],\varphi(X_{j}^{\prime})\subset Y_{\sigma(j)}^{\prime}$, and we denote the induced restricted map by $\varphi_{j}:X_{j}^{\prime}\to Y_{\sigma(j)}^{\prime}$ (Claim 2 is used here). Thus for each $j\in[m]$ there exists $f_{j}:X_{j}\to Y_{\sigma(j)}$ associated with $\varphi_{j}$ by Claim 1 after shrinking $S$ finitely many times.
2. for each pair $\alpha,\beta\in[n]$, there exists a cover of $Y_{\alpha\beta}$ by finite open subsets $\{Y_{\alpha\beta k}\}_{k\in[n_{\alpha\beta}]}$ distinguished in both $Y_{\alpha}$ and $Y_{\beta}$;
3. for each pair $\alpha,\beta\in[m]$, there exists a cover of $X_{\alpha\beta}$ by finite open subsets $\{X_{\alpha\beta k}\}_{k\in[m_{\alpha\beta}]}$ distinguished in both $X_{\alpha}$ and $X_{\beta}$, and a function $\sigma_{\alpha\beta}:[m_{\alpha\beta}]\to[n_{\sigma(\alpha)\sigma(\beta)}]$ s.t. for each $k\in[m_{\alpha\beta}]$, $\varphi(X_{\alpha\beta k}^{\prime})\subset Y_{\sigma(\alpha)\sigma(\beta)\sigma_{\alpha\beta}(k)}^{\prime}$. By Claim 3, we can see that, $f_{\alpha}(X_{\alpha\beta k})\subset Y_{\sigma(\alpha)\sigma(\beta)\sigma_{\alpha\beta}(k)}$ and $f_{\beta}(X_{\alpha\beta k})\subset Y_{\sigma(\alpha)\sigma(\beta)\sigma_{\alpha\beta}(k)}$ after shrinking $S$ finitely many times. So them induce two maps $X_{\alpha\beta k}\to Y_{\sigma(\alpha)\sigma(\beta)\sigma_{\alpha\beta}(k)}$ agreeing over $\Spec\mathcal{O}_{S,s}$, thus we can assume they are the same by shrinking $S$ (Claim 1).

Note that everything is Noetherian so quasi-compact. Clearly $Y$ admits a finite open affine covering $\{Y_{i}=\Spec A_{i}\}_{i\in[n]}$ s.t. for each pair $\alpha,\beta\in[n]$, $Y_{\alpha\beta}$ is covered by finite open subsets $\{Y_{\alpha\beta k}\}_{k\in[n_{\alpha\beta}]}$ distinguished in both $Y_{\alpha}$ and $Y_{\beta}$. Similarly $X$ admits a finite open affine covering $\{Z_{j}=\Spec B_{j}\}_{j\in[p]}$. Fix $j\in[p]$, $Z_{j}^{\prime}$ is covered by $\{\varphi^{-1}(Y_{i}^{\prime})\cap Z_{j}^{\prime}\}_{i\in[n]}$, and each of them is again covered by finitely many distinguished affine opens of $Z_{j}^{\prime}$, say $Z_{j}^{\prime}=\bigcup_{i,k}U_{ijk}$. By Claim 2 (i) and (ii), each $U_{ijk}$ is of the form $V_{ijk}^{\prime}$ where $V_{ijk}$ is distinguished affine open in $Z_{j}$ and we have $Z_{j}=\bigcup_{i,k}V_{ijk}$ after shrinking $S$. Clearly $\varphi(V_{ijk}^{\prime})\subset Y_{i}^{\prime}$. Let $\{X_{j}\}_{j\in[m]}=\{V_{ijk}\}_{i,j,k}$. Then we have shown part (1) and part (2).

Fix a pair $\alpha,\beta\in[m]$. Note that $\{Y_{\sigma(\alpha)\sigma(\beta)k}\}_{k\in[n_{\sigma(\alpha)\sigma(\beta)}]}$ covers $Y_{\sigma(\alpha)\sigma(\beta)}$. Also $X_{\alpha\beta}$ admits a finite open covering $\{\widetilde{X}_{\alpha\beta k}\}_{k\in[q_{\alpha\beta}]}$ distinguished in both $X_{\alpha}$ and $X_{\beta}$. Similarly, fix $k\in[q_{\alpha\beta}]$, $\widetilde{X}_{\alpha\beta k}^{\prime}$ is covered by $\{\varphi^{-1}(Y_{\sigma(\alpha)\sigma(\beta)k^{\prime}}^{\prime})\cap\widetilde{X}_{\alpha\beta k}^{\prime}\}_{k^{\prime}\in[n_{\sigma(\alpha)\sigma(\beta)}]}$, and each of them is again covered by finitely many distinguished affine opens of $\widetilde{X}_{\alpha\beta k}^{\prime}$, say $\widetilde{X}_{\alpha\beta k}^{\prime}=\bigcup_{k^{\prime},t}U_{\alpha\beta kk^{\prime}t}$. By Claim 2 (i) and (ii), each $U_{\alpha\beta kk^{\prime}t}$ is of the form $V_{\alpha\beta kk^{\prime}t}^{\prime}$ where $V_{\alpha\beta kk^{\prime}t}$ is distinguished affine open in $\widetilde{X}_{\alpha\beta k}$ and we have $\widetilde{X}_{\alpha\beta k}=\bigcup_{k^{\prime},t}V_{\alpha\beta kk^{\prime}t}$ after shrinking $S$. Clearly $\varphi(V_{\alpha\beta kk^{\prime}t})\subset Y_{\sigma(\alpha)\sigma(\beta)k^{\prime}}.$Let $\{X_{\alpha\beta k}\}_{k\in[m_{\alpha\beta}]}=\{V_{\alpha\beta kk^{\prime}t}\}_{k,k^{\prime},t}$. This completes part (3).

It follows that for all $\alpha,\beta\in[m]$, for all $k\in[m_{\alpha\beta}]$, we have $f_{\alpha}|_{X_{\alpha\beta k}}=f_{\beta}|_{X_{\alpha\beta k}}$ (assuming the target is $Y$). Thus $f_{\alpha}|_{X_{\alpha\beta}}=f_{\beta}|_{X_{\alpha\beta}}$ (assuming the target is $Y$). Hence $\{f_{\alpha}\}_{\alpha\in[m]}$ (target $Y$) glues to a desired morphism $f:X\to Y$.

Next we show the part of about lifting an isomorphism. Note that we can assume $S$ is affine. By the same method in the proof of Claim 1, it suffices to show that if $(D(g_{1}),f_{1}:X_{D(g_{1})}\to Y_{D(g_{1})})$ and $(D(g_{2}),f_{2}:X_{D(g_{2})}\to Y_{D(g_{2})})$ are two pairs satisfying the conditions in (a), then there exists $h\in R\backslash\mathfrak{p}$ s.t. $f_{1}=f_{2}$ over $D(g_{1}g_{2}h)$.

By shrinking $S$ to $D(g_{1}g_{2})$, we can assume $S=D(g_{1})=D(g_{2})$ and we have $f_{1},f_{2}:X\to Y$ inducing $\varphi$. Again $Y$ admits a finite open affine covering $\{Y_{i}=\Spec A_{i}\}_{i\in[n]}$. For each $i\in[n]$, $\varphi^{-1}(Y_{i}^{\prime})=f_{1}^{-1}(Y_{i})^{\prime}=f_{2}^{-1}(Y_{i})^{\prime}=(f_{1}^{-1}(Y_{i})\cap f_{2}^{-1}(Y_{i}))^{\prime}$. By Claim 2 (iv), we have $f_{1}^{-1}(Y_{i})=f_{1}^{-1}(Y_{i})\cap f_{2}^{-1}(Y_{i})=f_{2}^{-1}(Y_{i})$ after shrinking $S$. Denote $T_{i}:=f_{1}^{-1}(Y_{i})=f_{2}^{-1}(Y_{i})$. It suffices to show that we can shrink $S$ finitely many times so that $f_{1}|_{T_{i}}=f_{2}|_{T_{i}}$ (with target $Y_{i}$) for each $i$. So we may assume $Y$ is affine. Let $\{X_{i}\}_{i\in[m]}$ be a finite open affine covering of $X$. By Claim 1, for each $i\in[m]$, we have that $f_{1}|_{X_{i}}=f_{2}|_{X_{i}}$ by shrinking $S$ to a distinguished open subset. Then we just do this finitely many times so that $f_{1}=f_{2}$. The result follows. $\square$.

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Claim 2: Let $R$ be a Noetherian ring and $\mathfrak{p}\in\Spec R$, $A$ be a finitely generated $R$-algebra. Then:

• (i) Any distinguished affine open subset of $\Spec A_{\mathfrak{p}}$ is of the form $\Spec(A_{a})_{\mathfrak{p}}$ for some $a\in A$;

• (ii) If $\Spec A_{\mathfrak{p}}$ is covered by $D(a_{i})_{\mathfrak{p}}:=\Spec(A_{a_{i}})_{\mathfrak{p}}$ for $1\leq i\leq n$, then there exists $f\in R\backslash\mathfrak{p}$ s.t. $\Spec A_{f}$ is covered by $D(a_{i})_{f}:=\Spec(A_{a_{i}})_{f}$ for $1\leq i\leq n$.

• (iii) If there exist an open subset $U\subset\Spec A$ s.t. $U_{\mathfrak{p}}:=U\times_{\Spec R}\Spec R_{\mathfrak{p}}=\Spec A_{\mathfrak{p}}$, then there exists $f\in R\backslash\mathfrak{p}$ s.t. $U_{f}:=U\times_{\Spec R}\Spec R_{f}=\Spec A_{f}$.

• (iv) If $X$ is of finite type over $R$ and there exists an open subset $U\subset X$ s.t. $U_{\mathfrak{p}}=X_{\mathfrak{p}}$, then there exists $f\in R\backslash\mathfrak{p}$ s.t. $U_{f}=X_{f}$.

Proof: (i) Let $D(\frac{a}{r})\subset\Spec A_{\mathfrak{p}}$ with $a\in A,r\in R\backslash\mathfrak{p}$. Since $r$ is a unit in $A_{\mathfrak{p}}$, we have $(A_{\mathfrak{p}})_{\frac{a}{r}}=(A_{\mathfrak{p}})_{a}=(A_{a})_{\mathfrak{p}}$. (ii) Note that $\Spec(A_{a_{i}})_{\mathfrak{p}}=\Spec(A_{\mathfrak{p}})_{a_{i}}$. There exists $\frac{b_{i}}{r_{i}}\in A_{\mathfrak{p}}$ s.t. $$ 1=\sum_{i}\frac{b_{i}}{r_{i}}\cdot a_{i}\in A_{\mathfrak{p}}. $$ Then we can lift this relation to $A_{f}$ for some $f\in R\backslash\mathfrak{p}$. The result follows. (iii) follows from (i) and (ii) and the fact that $U$ is quasi-compact. (iv) Since $X$ is quasi-compact, $X$ admits a finite open affine covering $\{X_{i}\}_{i\in[m]}$. For each $i\in[m]$, denote $U_{i}:=X_{i}\cap U$. Then $(U_{i})_{\mathfrak{p}}=(X_{i})_{\mathfrak{p}}\cap U_{\mathfrak{p}}=(X_{i})_{\mathfrak{p}}\cap X_{\mathfrak{p}}=(X_{i})_{\mathfrak{p}}$. By (iii), there exists $f_{i}\in R\backslash\mathfrak{p}$ s.t. $(U_{i})_{f_{i}}=(X_{i})_{f_{i}}$. Let $f=\prod_{i}f_{i}$, then $U_{f}=X_{f}$. $\square$

Claim 3: Let $R$ be a ring and $\mathfrak{p}\in\Spec R$, $A,B$ be two $R$-algebras. Let $a\in A,b\in B$, and $\varphi:A\to B$ be a map of $R$-algebra s.t. it induces a map $\psi:(A_{\mathfrak{p}})_{a}\to(B_{\mathfrak{p}})_{b}$ (i.e. the image of $a$ is invertible in $(B_{\mathfrak{p}})_{b}$). Then there exists $f\in R\backslash\mathfrak{p}$ s.t. $\varphi$ induces a map $\phi:(A_{f})_{a}\to(B_{f})_{b}$ (i.e. the image of $a$ is invertible in $(B_{f})_{b}$).

Proof: Say $\psi(a)\cdot\frac{b^{\prime}}{b^{m}\cdot f^{\prime}}=\varphi(a)\cdot\frac{b^{\prime}}{b^{m}\cdot f^{\prime}}=1\in(B_{\mathfrak{p}})_{b}$ with $m\geq1,f^{\prime}\in R\backslash\mathfrak{p}$. Clearly there exist $f\in R\backslash\mathfrak{p}$ s.t. the equality holds in $(B_{f})_{b}$. The result follows.$\square$

Answer 4

(This answer is based on what previously discussed) In the comments it is remarked we have lots of maps $\varphi_i: Spec\ A_i\otimes R_r \rightarrow Y\times Spec\ R_r$. If we localize we get the map $Spec\ A_i\otimes R_p\rightarrow Spec\ B_j\otimes R_p\hookrightarrow Y\times Spec\ R_p$ given in the hypothesis.

Lemma Let $\varphi: X\times Spec\ R_p \rightarrow Y\times Spec\ R_p$ as in our hypothesis and $f,g:X\times Spec\ R_r \rightarrow Y\times Spec\ R_r$ two maps whose localizations coincide with $\varphi$. Then $f$ and $g$ are the same map (up to shrinking $Spec\ R_r$).

Proof As it was pointed out in the first answer there exist affine opens $Spec\ A_i$ and $Spec\ B_j$ resp. of $X$ and $Y$ such that $\forall\ i$ $Spec\ A_i\otimes R_p$ goes into $Spec\ B_j\otimes R_p$ through $\varphi$ for some $j$. Looking at the rings map we get a map $B_j\otimes R_p\rightarrow A_i\otimes R_p$ and so maps $\hat{f},\hat{g}: B_j\otimes R_r\rightarrow B_j\otimes R_p\rightarrow A_i\otimes R_p$ (this just considering the commutative diagram given by the fibre products). Up to further localizations one can assume both maps factorize through $B_j\otimes R_r\rightarrow A_i\otimes R_r\rightarrow A_i\otimes R_p$ so that we can assume $f,g:Spec\ A_i\otimes R_r \rightarrow Spec\ B_j\otimes R_r \hookrightarrow Y\times Spec\ R_r$. Restricting again $Spec\ R_r$ (and working with the ring maps) we can adfirm $f=g$ on $Spec\ A_i\otimes R_r$. You get the thesis after finitely many of such operations.

How to use this lemma? We have maps $\varphi_i: Spec\ A_i\otimes R_r \rightarrow Y\times Spec\ R_r$ and if we consider the intersections $V_{ijr}:=Spec\ A_i\otimes R_r \cap Spec\ A_j\otimes R_r$ we obtain two maps with same localization from $V_{ijr}$ to $Y\times Spec\ R_r$. Using the lemma we can make them coicide and glue them. I think one can use the lemma above for the statement about the isomorphism too.