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?
Best Answer
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.