# Gluing affine schemes

Let $Y$ be a scheme, and $\mathcal{A}$ be a sheaf of $\mathcal{O}_Y$-algebras. Such that $\mathcal{A}$ is quasi-coherent.

For every affine open set $V$ in $Y$, we have ring morphisms $\mathcal{O}_Y(V) \to \mathcal{A}(V)$ which induces $\text{spec} \mathcal{A}(V) \to \text{spec} \mathcal{O}_Y(V)$. Now $\text{spec} \mathcal{O}_Y(V)\simeq V$ naturally. Thus, we have a collection of natural scheme morphisms $f_V: \text{spec} \mathcal{A}(V) \to V$.

Construct $X$ to be a topological space obtained gluing each of the $f_V$ maps. Therefore, one has continuous map $\lambda_V: \text{spec} \mathcal{A}(V) \to X$, and $f:X\to Y$, such that $f_V = f\circ \lambda_V$.

Denote by $X_V$ to be the image of $\lambda_V$, this is an open subset of $X$ homeomorphic to $\text{spec} \mathcal{A}(V)$. One can then push the structure sheaf and construct $(X_V,F_V)$ which is a scheme.

The open sets $X_V$ cover $X$, and over each one there is a sheaf $F_V$. Can one glue these sheaves $F_V$ together here?

In order to glue, one requires morphisms $\varphi_{VU}:F_V|_{X_V\cap X_U} \to F_U|_{X_V\cap X_U}$, which satisfy the co-cycle condition.

Let $W$ be an open set in $X_V\cap X_U$, then $\lambda^{-1}_VW$ is open in $\text{spec} \mathcal{A}(V)$ and $\lambda^{-1}_UW$ is open in $\text{spec} \mathcal{A}(U)$. These two schemes $\lambda^{-1}_VW$ and $\lambda^{-1}_UW$ are isomorphic.

But how does one construct an isomorphism between them that would carry over and give us the desired morphism between $F_V$ and $F_U$ satisfying the co-cycle condition?

• Probably you want to add a quasi-coherence hypothesis, because it is not clear that you can glue the morphisms $f_V$ to obtain a topological space $X$. Consider the case that $Y$ is $\mathbb{A}^1_k = \text{Spec}\ k[t]$ and $\mathcal{A}$ is the $\mathcal{O}_Y$-algebra such that $\mathcal{A}(V)$ equals $\mathcal{O}_Y(V)$ for every open neighborhood $V$ of $t=0$, yet $\mathcal{A}(W)$ equals the zero ring for every open $W$ that does not contain $t=0$. For an open neighborhood $V$ of $0$ and for $W=V\setminus \{0\}$, it is not true that $f_V^{-1}(W)$ equals $\text{Spec}\ \mathcal{A}(W)$. – Jason Starr Jan 6 '16 at 0:25
• Don't you just use the fact that $A$ is a sheaf? Isn't this just the standard construction of relative spec? Isn't there some standard reference for this? – eric Jan 6 '16 at 7:43
• @eric: As I explained above, you certainly need to assume that $\mathcal{A}$ is quasi-coherent to get precisely what the OP asks. Fixing that hypothesis: yes, this is a standard construction, and yes there is a standard reference for this. – Jason Starr Jan 6 '16 at 8:50
• EGA II.1.3? You could also define this without gluing: take the underlying topological space to be the set of quasicoherent sheaves of ideals which are prime (which means that sections over each open are prime or the unit ideal.) Now proceed exactly as in the usual case- defining standard opens and the structure sheaf, etc. – Dylan Wilson Jan 6 '16 at 12:15
• Watch out Dylan, that section of Eisenbud-Harris is plagued with errors! "Prime sheaves" is not a good concept, e.g., the set of prime ideal sheaves of the structure sheaf is not naturally identified with the points of the scheme itself. – Dave Anderson Jan 10 '16 at 17:13

## 2 Answers

I just want to correct one mistake in my counterexample, and I want to make another philosophical point. First, you do need some hypothesis such as quasi-coherence of $\mathcal{A}$. The counterexample I wrote is wrong; here is a correction. Let $Y$ be $\mathbb{A}^1_k = \text{Spec}\ k[t]$. Let $0$ be the closed point with corresponding maximal ideal $\langle t \rangle \subset k[t]$. Let $\mathcal{A}$ be the $\mathcal{O}_Y$-module such that $\mathcal{A}(U)$ is $\{0\} = k[t]/\langle 1 \rangle$ if $U$ does not contain $0$, and such that $\mathcal{A}(U) = k(t)$ as a $k[t]$-algebra if $U$ does contain $0$. For open subsets $V\subset U$, the restriction homomorphism $\mathcal{A}(U)\to\mathcal{A}(V)$ is either zero, if $V$ does not contain $0$, or the identity on $k(t)$ if $V$ does contain $0$. It is straightforward to check that this is a sheaf (this is the mistake in my previous example). For every open affine $U\subset Y$, there is a natural morphism $$f_U:\text{Spec}\ \mathcal{A}(U) \to U,$$ and for every pair of open affines $V\subset U \subset Y$, there is a natural commutative diagram, $$\begin{array}{ccc} \text{Spec}\ \mathcal{A}(V) & \xrightarrow{f_V} & V \\ \downarrow & & \downarrow \\ \text{Spec}\ \mathcal{A}(U) & \xrightarrow{f_U} & U \end{array}.$$ However, this commutative diagram is not a fiber product diagram: if $U$ contains $0$ but $V$ does not, the fiber product is $\text{Spec}\ k(t)$, but $\mathcal{A}(V)$ is the zero ring. Therefore, there is no morphism $f:X\to Y$ and collection of isomorphisms $$\phi_U:f^{-1}(U) \xrightarrow{\cong} \text{Spec}\ \mathcal{A}(U),$$ as schemes over $U$. If there were, then $\text{Spec}\ \mathcal{A}(V)$ as above would be $f^{-1}(V)$, and this is the fiber product of $f^{-1}(U)\to U$ and $V\to U$.

Now here is the philosophical point. One way to "calibrate" the gluing is to write down a universal property satisfied by the glued object and that is compatible with restricting to open subsets. In this case, the universal property of $f:X\to Y$ is that there is a homomorphism of $\mathcal{O}_X$-algebras, $s:f^*\mathcal{A}\to \mathcal{O}_X$, and this is universal: for every $g:Z\to Y$ and homomorphism of $\mathcal{O}_Z$-algebras, $t:g^*\mathcal{A}\to \mathcal{O}_Z$, there exists a unique morphism $h:Z\to X$ such that $f\circ h$ equals $g$ and such that $t$ equals $h^*s:h^*f^*\mathcal{A}\to h^*\mathcal{O}_X$. In the quasi-coherent case, where locally $\mathcal{A}_U = \widetilde{A_U}$ for an $\mathcal{O}_Y(U)$-algebra $A_U$, it is straightforward to check that locally $\text{Spec}\ \mathcal{A}(U) \to U$ satisfies the universal property. Then, for open affines $U$ and $V$, the corresponding universal properties give an isomorphism of $f_U^{-1}(U\cap V)$ and $f_V^{-1}(U\cap V)$ as schemes over $U\cap V$. Finally, the uniqueness part of the universal property implies the cocycle condition for these isomorphisms.

I am going to follow the method suggested by Jason Starr. Unfortunately, I reach a dead-end in the proof, hopefully one of you can help me fix it.

Note: I modified the universal property that we want. Because I want to be consistent with Hartshorne Exercise 2.5.17.

Let $(Y,\mathcal{O}_Y)$ be a scheme with $\mathscr{A}$ a quasi-coherent sheaf of $\mathcal{O}_Y$-algebras. We denote by $\text{spec } \mathscr{A}$ to be a scheme $X$, for which we have a scheme morphism $f:X\to Y$, and a collection of isomorphisms $\theta_V:f^{-1}V \to \text{spec }\mathscr{A}(V)$ for each open affine $V$ in $Y$, such that if $V\subseteq U$ then the following diagram commutes, $$\require{AMScd} \begin{CD} f^{-1} V @>>> f^{-1} U \\ @V\theta_{V} VV @VV \theta_U V \\ \text{spec } \mathscr{A}(V) @>>> \text{spec } \mathscr{A}(U) \end{CD}$$ Furthermore, we require $X$ to have the following universal property. If $Z$ is any scheme for which we have a morphism $g:Z\to Y$, and a natural collection of isomorphisms $\zeta_V:g^{-1}V\to \text{spec } \mathscr{A}(V)$, then there exists a unique morphism $h:Z\to X$ such that $g = f\circ h$ and such that $\zeta_V = \theta_V\circ h_V$ (in the second equation, $h$ is being restricted to $h_V:g^{-1}V \to f^{-1}V$).

Clearly, by abstract non-sense if $X$ exists it is unique, so the question remains that of existence. We first focus on the case when $Y$ is an affine scheme, so $Y = (\text{spec } A, \widetilde{A})$ for some ring $A$. The quasicoherent sheaf $\mathscr{A}$ is then of the form $\mathscr{A} = \widetilde{B}$ for some algebra $B$ over $A$, say given the ring homomorphism $f':A\to B$. Thus, one obtains a scheme morphism $f:\text{spec } B \to \text{spec } A$, and we claim that $\text{spec } B$ will satisfy the required universal property above. One now has to define a natural family of isomorphisms $\theta_V: f^{-1}V\to \text{spec } \mathscr{A}(V)$ for each affine open set $V$ in $Y=\text{spec } A$. First say $V = D(a)$, a basic open set in $Y$, for some $a\in A$. Then $f^{-1}V = D(f'(a))$ a basic open set in $X$. Now $\mathscr{A}(V) = \widetilde{B}(D(a)) = B_a$. A simple algebra exercise will show that the localization $B_a = B_{f'(a)}$. Hence, we need a natural isomorphism $\theta_{D(a)}: D(f'(a)) \to \text{spec } B_{f'(a)}$, which we certainly have (from applying spec to the localization map $B\to B_{f'(a)}$, cf. Exercise 2.1.).

Here is the nice thing about our family $\theta_V$, $V$ basic open, of isomorphisms so far. They are compatible with restriction. That is to say, if $V\subseteq U$ are basic open sets, then the required commutative diagram will be satisfied. But how do we construct the isomorphism $\theta_V$ when $V$ is a more complicated affine set? One invokes the following Lemma which is left unproved as an exercise.

Lemma. Let $f:X\to Y$ be a continuous map of schemes. Let $V_i$ be a collection of open sets which covers a scheme $Y$. For each $i$ we have a scheme (iso)morphism $(f,\theta_i):f^{-1}V_i \to V_i$, where the topological map is given by $f$, and $\theta_i$ is a morphism of structure sheaves . These (iso)morphisms satisfy: if $V_i\subseteq V_j$ then the dia. com., $$\begin{CD} \mathcal{O}_Y(V_j) @>>> \mathcal{O}_Y(V_i) \\ @V\theta_j VV @VV \theta_i V \\ \mathcal{O}_X(f^{-1}V_j) @>>> \mathcal{O}_X(f^{-1}V_i) \end{CD}$$ Under these hypothesis there exists an (iso)morphism $(f,\theta):X\to Y$. Furthermore, this (iso)morphism is also compatible with restriction.

Now to see why $X$ satisfies the required universal property. Let $h:Z\to Y$ be a scheme for which we have a natural collection of isomorphisms $\zeta_V:h^{-1}V\to \text{spec } \mathscr{A}(V)$. By choosing $V=Y$ we obtain an isomorphism $\zeta_Y:Z\to \text{spec } B$. Call $h=\zeta_Y$. We check that $h$ satisfies $g = f\circ h$ and $\zeta_V = \theta_V\circ h_V$. By choosing $V=Y$, $h_Y = h$ and $\theta_Y = \text{id}$, so we get $\zeta_Y = \text{id}\circ h$ which is true since we defined $h = \zeta_Y$. Therefore, one has to only verify the condition $\zeta_V = \theta_V\circ h_V$.

By using $V\subseteq Y$, where $V$ is any affine open in $Y$, we get a com. dia.,
$$\begin{CD} g^{-1} V @>i>> Z \\ @V\zeta_{V} VV @V V h V \\ \text{spec }\mathscr{A}(V) @>>> X \\ @V\theta_{V}^{-1} VV @VV\text{id}V \\ f^{-1}V @>>j> X \end{CD}$$ From this diagram we see that $\theta_V^{-1}\circ \zeta_V:g^{-1}V\to f^{-1}V$ and $h_V:g^{-1}V\to f^{-1}V$ are equal. Thus, we obtain $\zeta_V = \theta_V\circ h_V$ as required.

Thus, $\text{spec } \mathscr{A}$ exists in the case when $Y$ is an affine scheme and it is given by $\text{spec } B$. For the next part of the proof we show that if $Y$ is a scheme with a quasi-coherent sheaf $\mathscr{A}$ of $\mathcal{O}_Y$-algebras, and $U$ is an open subset of $Y$, then $\text{spec } \mathscr{A}|_U$ exists when we restrict the sheaf $\mathscr{A}$ to the scheme $U$. In fact, if $f:X\to Y$ is our universal map where $X=\text{spec }\mathscr{A}$, then $f^{-1}U = \text{spec } \mathscr{A}|_U$. Since $X = \text{spec } \mathscr{A}$, for each affine open set $V$ in $Y$, we have a natural collection of isomorphisms $\theta_V: f^{-1}V\to \text{spec } \mathscr{A}(V)$. By choosing those affine open sets contained in $U$, the collection $\{ \theta_V \}|_{V\subseteq U}$ is a natural collection, for the scheme $U$ with quasicoherent $\mathcal{O}_Y|_U$-algebra given by $\mathscr{A}|_U$.

One has a scheme morphism $f^{-1}U\to U$ given by restriction called $f_U$, and the natural collection $\{ \theta_V \}|_{V\subseteq U}$. We will show that these two satisfy the universal property for $\text{spec } \mathscr{A}|_U$. We suppose that there is a scheme $S$ together with a morphism $g:S\to U$ and a natural collection of isomorphism $\sigma_V:g^{-1}V\to \text{spec } \mathscr{A}(V)$, for every open affine $V$ contained in $U$. We seek to exhibit a morphism $h:S\to f^{-1}U$, such that $g = f_U\circ h$ and $\sigma_V = \theta_V \circ h_V$.

Here are the problems. (i) How do I complete the proof that $f^{-1}U$ is the relative spec of $\mathscr{A}|_U$? (ii) It seems in my approach I require the assumption that $Y$ is separated?