## 26.14 Glueing schemes

Let $I$ be a set. For each $i \in I$ let $(X_ i, \mathcal{O}_ i)$ be a locally ringed space. (Actually the construction that follows works equally well for ringed spaces.) For each pair $i, j \in I$ let $U_{ij} \subset X_ i$ be an open subspace. For each pair $i, j \in I$, let

\[ \varphi _{ij} : U_{ij} \to U_{ji} \]

be an isomorphism of locally ringed spaces. For convenience we assume that $U_{ii} = X_ i$ and $\varphi _{ii} = \text{id}_{X_ i}$. For each triple $i, j, k \in I$ assume that

we have $\varphi _{ij}^{-1}(U_{ji} \cap U_{jk}) = U_{ij} \cap U_{ik}$, and

the diagram

\[ \xymatrix{ U_{ij} \cap U_{ik} \ar[rr]_{\varphi _{ik}} \ar[rd]_{\varphi _{ij}} & & U_{ki} \cap U_{kj} \\ & U_{ji} \cap U_{jk} \ar[ru]_{\varphi _{jk}} } \]

is commutative.

Let us call a collection $(I, (X_ i)_{i\in I}, (U_{ij})_{i, j\in I}, (\varphi _{ij})_{i, j\in I})$ satisfying the conditions above a glueing data.

slogan
Lemma 26.14.1. Given any glueing data of locally ringed spaces there exists a locally ringed space $X$ and open subspaces $U_ i \subset X$ together with isomorphisms $\varphi _ i : X_ i \to U_ i$ of locally ringed spaces such that

$\varphi _ i(U_{ij}) = U_ i \cap U_ j$, and

$\varphi _{ij} = \varphi _ j^{-1}|_{U_ i \cap U_ j} \circ \varphi _ i|_{U_{ij}}$.

The locally ringed space $X$ is characterized by the following mapping properties: Given a locally ringed space $Y$ we have

\begin{eqnarray*} \mathop{Mor}\nolimits (X, Y) & = & \{ (f_ i)_{i\in I} \mid f_ i : X_ i \to Y, \ f_ j \circ \varphi _{ij} = f_ i|_{U_{ij}}\} \\ f & \mapsto & (f|_{U_ i} \circ \varphi _ i)_{i \in I} \\ \mathop{Mor}\nolimits (Y, X) & = & \left\{ \begin{matrix} \text{open covering }Y = \bigcup \nolimits _{i \in I} V_ i\text{ and } (g_ i : V_ i \to X_ i)_{i \in I} \text{ such that}
\\ g_ i^{-1}(U_{ij}) = V_ i \cap V_ j \text{ and } g_ j|_{V_ i \cap V_ j} = \varphi _{ij} \circ g_ i|_{V_ i \cap V_ j}
\end{matrix} \right\} \\ g & \mapsto & V_ i = g^{-1}(U_ i), \ g_ i = \varphi _ i^{-1} \circ g|_{V_ i} \end{eqnarray*}

**Proof.**
We construct $X$ in stages. As a set we take

\[ X = (\coprod X_ i) / \sim . \]

Here given $x \in X_ i$ and $x' \in X_ j$ we say $x \sim x'$ if and only if $x \in U_{ij}$, $x' \in U_{ji}$ and $\varphi _{ij}(x) = x'$. This is an equivalence relation since if $x \in X_ i$, $x' \in X_ j$, $x'' \in X_ k$, and $x \sim x'$ and $x' \sim x''$, then $x' \in U_{ji} \cap U_{jk}$, hence by condition (1) of a glueing data also $x \in U_{ij} \cap U_{ik}$ and $x'' \in U_{ki} \cap U_{kj}$ and by condition (2) we see that $\varphi _{ik}(x) = x''$. (Reflexivity and symmetry follows from our assumptions that $U_{ii} = X_ i$ and $\varphi _{ii} = \text{id}_{X_ i}$.) Denote $\varphi _ i : X_ i \to X$ the natural maps. Denote $U_ i = \varphi _ i(X_ i) \subset X$. Note that $\varphi _ i : X_ i \to U_ i$ is a bijection.

The topology on $X$ is defined by the rule that $U \subset X$ is open if and only if $\varphi _ i^{-1}(U)$ is open for all $i$. We leave it to the reader to verify that this does indeed define a topology. Note that in particular $U_ i$ is open since $\varphi _ j^{-1}(U_ i) = U_{ji}$ which is open in $X_ j$ for all $j$. Moreover, for any open set $W \subset X_ i$ the image $\varphi _ i(W) \subset U_ i$ is open because $\varphi _ j^{-1}(\varphi _ i(W)) = \varphi _{ji}^{-1}(W \cap U_{ij})$. Therefore $\varphi _ i : X_ i \to U_ i$ is a homeomorphism.

To obtain a locally ringed space we have to construct the sheaf of rings $\mathcal{O}_ X$. We do this by glueing the sheaves of rings $\mathcal{O}_{U_ i} := \varphi _{i, *} \mathcal{O}_ i$. Namely, in the commutative diagram

\[ \xymatrix{ U_{ij} \ar[rr]_{\varphi _{ij}} \ar[rd]_{\varphi _ i|_{U_{ij}}} & & U_{ji} \ar[ld]^{\varphi _ j|_{U_{ji}}} \\ & U_ i \cap U_ j & } \]

the arrow on top is an isomorphism of ringed spaces, and hence we get unique isomorphisms of sheaves of rings

\[ \mathcal{O}_{U_ i}|_{U_ i \cap U_ j} \longrightarrow \mathcal{O}_{U_ j}|_{U_ i \cap U_ j}. \]

These satisfy a cocycle condition as in Sheaves, Section 6.33. By the results of that section we obtain a sheaf of rings $\mathcal{O}_ X$ on $X$ such that $\mathcal{O}_ X|_{U_ i}$ is isomorphic to $\mathcal{O}_{U_ i}$ compatibly with the glueing maps displayed above. In particular $(X, \mathcal{O}_ X)$ is a locally ringed space since the stalks of $\mathcal{O}_ X$ are equal to the stalks of $\mathcal{O}_ i$ at corresponding points.

The proof of the mapping properties is omitted.
$\square$

slogan
Lemma 26.14.2. In Lemma 26.14.1 above, assume that all $X_ i$ are schemes. Then the resulting locally ringed space $X$ is a scheme.

**Proof.**
This is clear since each of the $U_ i$ is a scheme and hence every $x \in X$ has an affine neighbourhood.
$\square$

It is customary to think of $X_ i$ as an open subspace of $X$ via the isomorphisms $\varphi _ i$. We will do this in the next two examples.

Example 26.14.3 (Affine space with zero doubled). Let $k$ be a field. Let $n \geq 1$. Let $X_1 = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$, let $X_2 = \mathop{\mathrm{Spec}}(k[y_1, \ldots , y_ n])$. Let $0_1 \in X_1$ be the point corresponding to the maximal ideal $(x_1, \ldots , x_ n) \subset k[x_1, \ldots , x_ n]$. Let $0_2 \in X_2$ be the point corresponding to the maximal ideal $(y_1, \ldots , y_ n) \subset k[y_1, \ldots , y_ n]$. Let $U_{12} = X_1 \setminus \{ 0_1\} $ and let $U_{21} = X_2 \setminus \{ 0_2\} $. Let $\varphi _{12} : U_{12} \to U_{21}$ be the isomorphism coming from the isomorphism of $k$-algebras $k[y_1, \ldots , y_ n] \to k[x_1, \ldots , x_ n]$ mapping $y_ i$ to $x_ i$ (which induces $X_1 \cong X_2$ mapping $0_1$ to $0_2$). Let $X$ be the scheme obtained from the glueing data $(X_1, X_2, U_{12}, U_{21}, \varphi _{12}, \varphi _{21} = \varphi _{12}^{-1})$. Via the slight abuse of notation introduced above the example we think of $X_1, X_2 \subset X$ as open subschemes. There is a morphism $f : X \to \mathop{\mathrm{Spec}}(k[t_1, \ldots , t_ n])$ which on $X_1$ (resp. $X_2$) corresponds to $k$ algebra map $k[t_1, \ldots , t_ n] \to k[x_1, \ldots , x_ n]$ (resp. $k[t_1, \ldots , t_ n] \to k[y_1, \ldots , y_ n]$) mapping $t_ i$ to $x_ i$ (resp. $t_ i$ to $y_ i$). It is easy to see that this morphism identifies $k[t_1, \ldots , t_ n]$ with $\Gamma (X, \mathcal{O}_ X)$. Since $f(0_1) = f(0_2)$ we see that $X$ is not affine.

Note that $X_1$ and $X_2$ are affine opens of $X$. But, if $n = 2$, then $X_1 \cap X_2$ is the scheme described in Example 26.9.3 and hence not affine. Thus in general the intersection of affine opens of a scheme is not affine. (This fact holds more generally for any $n > 1$.)

Another curious feature of this example is the following. If $n > 1$ there are many irreducible closed subsets $T \subset X$ (take the closure of any non closed point in $X_1$ for example). But unless $T = \{ 0_1\} $, or $T = \{ 0_2\} $ we have $0_1 \in T \Leftrightarrow 0_2 \in T$. Proof omitted.

Example 26.14.4 (Projective line). Let $k$ be a field. Let $X_1 = \mathop{\mathrm{Spec}}(k[x])$, let $X_2 = \mathop{\mathrm{Spec}}(k[y])$. Let $0 \in X_1$ be the point corresponding to the maximal ideal $(x) \subset k[x]$. Let $\infty \in X_2$ be the point corresponding to the maximal ideal $(y) \subset k[y]$. Let $U_{12} = X_1 \setminus \{ 0\} = D(x) = \mathop{\mathrm{Spec}}(k[x, 1/x])$ and let $U_{21} = X_2 \setminus \{ \infty \} = D(y) = \mathop{\mathrm{Spec}}(k[y, 1/y])$. Let $\varphi _{12} : U_{12} \to U_{21}$ be the isomorphism coming from the isomorphism of $k$-algebras $k[y, 1/y] \to k[x, 1/x]$ mapping $y$ to $1/x$. Let $\mathbf{P}^1_ k$ be the scheme obtained from the glueing data $(X_1, X_2, U_{12}, U_{21}, \varphi _{12}, \varphi _{21} = \varphi _{12}^{-1})$. Via the slight abuse of notation introduced above the example we think of $X_ i \subset \mathbf{P}^1_ k$ as open subschemes. In this case we see that $\Gamma (\mathbf{P}^1_ k, \mathcal{O}) = k$ because the only polynomials $g(x)$ in $x$ such that $g(1/y)$ is also a polynomial in $y$ are constant polynomials. Since $\mathbf{P}^1_ k$ is infinite we see that $\mathbf{P}^1_ k$ is not affine.

We claim that there exists an affine open $U \subset \mathbf{P}^1_ k$ which contains both $0$ and $\infty $. Namely, let $U = \mathbf{P}^1_ k \setminus \{ 1\} $, where $1$ is the point of $X_1$ corresponding to the maximal ideal $(x - 1)$ and also the point of $X_2$ corresponding to the maximal ideal $(y - 1)$. Then it is easy to see that $s = 1/(x - 1) = y/(1 - y) \in \Gamma (U, \mathcal{O}_ U)$. In fact you can show that $\Gamma (U, \mathcal{O}_ U)$ is equal to the polynomial ring $k[s]$ and that the corresponding morphism $U \to \mathop{\mathrm{Spec}}(k[s])$ is an isomorphism of schemes. Details omitted.

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