Lemma 62.17.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. If $X$ is not a scheme, then there exists a closed subspace $Z \subset X$ such that $Z$ is not a scheme, but every proper closed subspace $Z' \subset Z$ is a scheme.

## 62.17 Obtaining schemes

A few more techniques to show an algebraic space is a scheme. The first is that we can show there is a minimal closed subspace which is not a scheme.

**Proof.**
We prove this by Zorn's lemma. Let $\mathcal{Z}$ be the set of closed subspaces $Z$ which are not schemes ordered by inclusion. By assumption $\mathcal{Z}$ contains $X$, hence is nonempty. If $Z_\alpha $ is a totally ordered subset of $\mathcal{Z}$, then $Z = \bigcap Z_\alpha $ is in $\mathcal{Z}$. Namely,

and the transition morphisms are affine. Thus we may apply Lemma 62.5.11 to see that if $Z$ were a scheme, then so would one of the $Z_\alpha $. (This works even if $Z = \emptyset $, but note that by Lemma 62.5.3 this cannot happen.) Thus $\mathcal{Z}$ has minimal elements by Zorn's lemma. $\square$

Now we can prove a little bit about these minimal non-schemes.

Lemma 62.17.2. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$. Assume that every proper closed subspace $Z \subset X$ is a scheme, but $X$ is not a scheme. Then $X$ is reduced and irreducible.

**Proof.**
We see that $X$ is reduced by Lemma 62.15.3. Choose closed subsets $T_1 \subset |X|$ and $T_2 \subset |X|$ such that $|X| = T_1 \cup T_2$. If $T_1$ and $T_2$ are proper closed subsets, then the corresponding reduced induced closed subspaces $Z_1, Z_2 \subset X$ (Properties of Spaces, Definition 58.12.6) are schemes and so is $Z = Z_1 \times _ X Z_2 = Z_1 \cap Z_2$ as a closed subscheme of either $Z_1$ or $Z_2$. Observe that the coproduct $Z_1 \amalg _ Z Z_2$ exists in the category of schemes, see More on Morphisms, Lemma 36.57.8. One way to proceed, is to show that $Z_1 \amalg _ Z Z_2$ is isomorphic to $X$, but we cannot use this here as the material on pushouts of algebraic spaces comes later in the theory. Instead we will use Lemma 62.15.1 to find an affine neighbourhood of every point. Namely, let $x \in |X|$. If $x \not\in Z_1$, then $x$ has a neighbourhood which is a scheme, namely, $X \setminus Z_1$. Similarly if $x \not\in Z_2$. If $x \in Z = Z_1 \cap Z_2$, then we choose an affine open $U \subset Z_1 \amalg _ Z Z_2$ containing $z$. Then $U_1 = Z_1 \cap U$ and $U_2 = Z_2 \cap U$ are affine opens whose intersections with $Z$ agree. Since $|Z_1| = T_1$ and $|Z_2| = T_2$ are closed subsets of $|X|$ which intersect in $|Z|$, we find an open $W \subset |X|$ with $W \cap T_1 = |U_1|$ and $W \cap T_2 = |U_2|$. Let $W$ denote the corresponding open subspace of $X$. Then $x \in |W|$ and the morphism $U_1 \amalg U_2 \to W$ is a surjective finite morphism whose source is an affine scheme. Thus $W$ is an affine scheme by Lemma 62.15.1.
$\square$

A key point in the following lemma is that we only need to check the condition in the images of points of $X$.

Lemma 62.17.3. Let $f: X \to S$ be a quasi-compact and quasi-separated morphism from an algebraic space to a scheme $S$. If for every $x \in |X|$ with image $s = f(x) \in S$ the algebraic space $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S,s})$ is a scheme, then $X$ is a scheme.

**Proof.**
Let $x \in |X|$. It suffices to find an open neighbourhood $U$ of $s = f(x)$ such that $X \times _ S U$ is a scheme. As $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is a scheme, then, since $\mathcal{O}_{S, s} = \mathop{\mathrm{colim}}\nolimits \mathcal{O}_ S(U)$ where the colimit is over affine open neighbourhoods of $s$ in $S$ we see that

By Lemma 62.5.11 we see that $X \times _ S U$ is a scheme for some $U$. $\square$

Instead of restricting to local rings as in Lemma 62.17.3, we can restrict to closed subschemes of the base.

Lemma 62.17.4. Let $\varphi : X \to \mathop{\mathrm{Spec}}(A)$ be a quasi-compact and quasi-separated morphism from an algebraic space to an affine scheme. If $X$ is not a scheme, then there exists an ideal $I \subset A$ such that the base change $X_{A/I}$ is not a scheme, but for every $I \subset I'$, $I \not= I'$ the base change $X_{A/I'}$ is a scheme.

**Proof.**
We prove this by Zorn's lemma. Let $\mathcal{I}$ be the set of ideals $I$ such that $X_{A/I}$ is not a scheme. By assumption $\mathcal{I}$ contains $(0)$. If $I_\alpha $ is a chain of ideals in $\mathcal{I}$, then $I = \bigcup I_\alpha $ is in $\mathcal{I}$. Namely, $A/I = \mathop{\mathrm{colim}}\nolimits A/I_\alpha $, hence

Thus we may apply Lemma 62.5.11 to see that if $X_{A/I}$ were a scheme, then so would be one of the $X_{A/I_\alpha }$. Thus $\mathcal{I}$ has maximal elements by Zorn's lemma. $\square$

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