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We know that the disjoint union of an infinite number of affine schemes is not an affine scheme since the underlying topological space is not quasi-compact.

But how do you show that the disjoint union of a finite number of affine schemes is an affine scheme? What are the global sections?

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The ring of global sections of the disjoint union (coproduct) is the product of the rings of global sections of the individual schemes: $$\bigcup^{\text{disjoint}}_{1\le i\le n} \text{Spec } R_i=\text{Spec }( R_1\times R_2\times \cdots\times R_n).$$

To explain this in more detail: Let $e_j$ be the element of $\hat R:=R_1\times\cdots\times R_n$ with $1$ in the $j$th position and $0$ elsewhere. Then, if $i\ne j$, any prime ideal $\hat P$ in $\hat R$ contains $e_i e_j=0$, so it contains either $e_i$ or $e_j$. Doing this for all pairs $\{i,j\}$ shows you that $\hat P$ must contain all of the $e_i$s except one, so it is of the form $R_1\times\cdots\times R_{i-1}\times P_i\times R_{i+1}\times\cdots\times R_n$, for some prime ideal $P_i$ of $R_i$. This shows that there is a natural correspondence between the points of $\text{Spec } \hat R$ and those of the disjoint union. Now, let $f_1\in R_1$, $\dots$, $f_n\in R_n$. Then the distinguished open set $D_{(f_1,\ldots,f_n)}$ of $\text{Spec } \hat R$ satisfies \begin{eqnarray*} D_{(f_1,\ldots,f_n)}&=&\{\hat P \subseteq \hat R\mid \hat P \text{ prime}, (f_1,\ldots,f_n)\notin \hat P\}\\ &=& \bigcup_{1\le i\le n} \{R_1\times\cdots\times P_i\times \cdots\times R_n\mid P_i \text{ prime}, f_i\notin P_i\}. \end{eqnarray*} Since these sets are a basis for the topology, if you then set $$ U_i:=D_{(0,\ldots,0,1,0,\ldots,0)}, \qquad i=1, \dots, n, \ \ \ \text{1 is at the }i\text{th position} $$ this shows that $$ \text{Spec } \hat R = U_1\cup\cdots\cup U_n $$ is a partition of the topological space of $\text{Spec } \hat R$ into $n$ disjoint clopen sets, and that the subset topology on each $U_i$ is identical with that of $\text{Spec } R_i$. Therefore, $\text{Spec } \hat R$ is the coproduct of the $\text{Spec } R_i$s in the category of topological spaces.

There is an isomorphism of localizations $${\hat R}_{(f_1,\ldots,f_n)} \cong (R_1)_{f_1}\times\cdots\times (R_n)_{f_n} $$ for which the square $$ \begin{array}{ccc} {\hat R}_{(f_1,\ldots,f_n)}&\cong&(R_1)_{f_1}\times\cdots\times(R_n)_{f_n}\\ \downarrow&\ &\downarrow\\ {\hat R}_{(g_1,\ldots,g_n)}&\cong&(R_1)_{g_1}\times\cdots\times(R_n)_{g_n} \end{array}$$ $$ (g_1 \in \sqrt{(f_1)}, \ \dots, \ g_n \in \sqrt{(f_n)}) $$ commutes. Since the structure sheaf of an affine scheme is assembled from these localizations, this can be used to show that there is an isomorphism $$ {\cal O}_{\text{Spec } \hat R}(U)\cong \prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(U)) $$ for all open sets $U\subseteq \text{Spec } \hat R$, where the map $\alpha_i: \text{Spec } R_i\to U_i$ is the map of topological spaces which sends $\text{Spec } R_i$ onto its homeomorphic image inside $\text{Spec } \hat R$. Also, if $V\subseteq U$, the diagram $$ \begin{array}{ccc} {\cal O}_{\text{Spec } \hat R}(U)&\cong&\prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(U))\\ \downarrow{}&&\downarrow{}\\ {\cal O}_{\text{Spec } \hat R}(V)&\cong&\prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(V)) \end{array} $$ commutes. This means that the stalk at each point in each $\text{Spec } R_i$ is isomorphic to the stalk at its image in $\text{Spec } \hat R$.

For each $i=1$, $\dots$, $n$, there is a morphism $\gamma_i: \text{Spec } R_i\to \text{Spec } \hat R$ which comes from the ring homomorphism projecting $\hat R$ onto $R_i$. To show that $\text{Spec } \hat R$ is the coproduct of the $\text{Spec } R_i$s in the category of schemes, you need to show that for any scheme $Z$ and morphisms $\psi_i: \text{Spec } R_i \to Z$, there is a unique morphism $\psi: \text{Spec } \hat R\to Z$ such that $\psi \circ \gamma_i=\psi_i$ for all $i$. Since $\text{Spec } \hat R$ is the disjoint union of the $\text{Spec } R_i$s as topological spaces, the action of $\psi$ on the topology is determined by letting it act as $\psi_i$ on the copy $U_i$ of $\text{Spec } R_i$ in $\text{Spec } \hat R$. Now for any open set $W$ in $Z$, the morphisms $\psi_i$ give you ring homomorphisms from ${\cal O}_Z(W)$ to ${\cal O}_{\text{Spec } R_i}(\psi_i^{-1}(W))$. Using the isomorphism between $\prod_i {\cal O}_{\text{Spec } R_i}(\psi_i^{-1}(W))$ and ${\cal O}_{\text{Spec } \hat R}(\psi^{-1}(W))$, these can be assembled into a ring homomorphism from ${\cal O}_Z(W)$ to ${\cal O}_{\text{Spec } \hat R}( \psi^{-1}(W))$. This produces the desired morphism $\psi$.

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  • $\begingroup$ Could you explain this in more detail? $\endgroup$
    – user45955
    Mar 5, 2013 at 3:36
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    $\begingroup$ I explained it in more detail above. $\endgroup$ Mar 6, 2013 at 3:26
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Suppoes $\mathrm Spec A_i$, $i=1,2$ are two affine spaces. Note that disjoint union is a co-product in category of set. So in this case, coproduct will be $\mathrm Spec (A_1\times A_2)$.

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