Lemma 71.10.5. Let $k$ be a field. Let $X$ be a quasi-separated algebraic space over $k$. If there exists a purely transcendental field extension $k \subset K$ such that $X_ K$ is a scheme, then $X$ is a scheme.

**Proof.**
Since every algebraic space is the union of its quasi-compact open subspaces, we may assume $X$ is quasi-compact (some details omitted). Recall (Fields, Definition 9.26.1) that the assumption on the extension $K/k$ signifies that $K$ is the fraction field of a polynomial ring (in possibly infinitely many variables) over $k$. Thus $K = \bigcup A$ is the union of subalgebras each of which is a localization of a finite polynomial algebra over $k$. By Limits of Spaces, Lemma 69.5.11 we see that $X_ A$ is a scheme for some $A$. Write

for some nonzero $f \in k[x_1, \ldots , x_ n]$.

If $k$ is infinite then we can finish the proof as follows: choose $a_1, \ldots , a_ n \in k$ with $f(a_1, \ldots , a_ n) \not= 0$. Then $(a_1, \ldots , a_ n)$ define an $k$-algebra map $A \to k$ mapping $x_ i$ to $a_ i$ and $1/f$ to $1/f(a_1, \ldots , a_ n)$. Thus the base change $X_ A \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(k) \cong X$ is a scheme as desired.

In this paragraph we finish the proof in case $k$ is finite. In this case we write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ with $X_ i$ of finite presentation over $k$ and with affine transition morphisms (Limits of Spaces, Lemma 69.10.2). Using Limits of Spaces, Lemma 69.5.11 we see that $X_{i, A}$ is a scheme for some $i$. Thus we may assume $X \to \mathop{\mathrm{Spec}}(k)$ is of finite presentation. Let $x \in |X|$ be a closed point. We may represent $x$ by a closed immersion $\mathop{\mathrm{Spec}}(\kappa ) \to X$ (Decent Spaces, Lemma 67.14.6). Then $\mathop{\mathrm{Spec}}(\kappa ) \to \mathop{\mathrm{Spec}}(k)$ is of finite type, hence $\kappa $ is a finite extension of $k$ (by the Hilbert Nullstellensatz, see Algebra, Theorem 10.34.1; some details omitted). Say $[\kappa : k] = d$. Choose an integer $n \gg 0$ prime to $d$ and let $k \subset k'$ be the extension of degree $n$. Then $k'/k$ is Galois with $G = \text{Aut}(k'/k)$ cyclic of order $n$. If $n$ is large enough there will be $k$-algebra homomorphism $A \to k'$ by the same reason as above. Then $X_{k'}$ is a scheme and $X = X_{k'}/G$ (Lemma 71.10.3). On the other hand, since $n$ and $d$ are relatively prime we see that

is the spectrum of a field. In other words, the fibre of $X_{k'} \to X$ over $x$ consists of a single point. Thus by Lemma 71.10.4 we see that $x$ is in the schematic locus of $X$ as desired. $\square$

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