If $X$ is a reduced algebraic space over a field, then it can happen that $X$ becomes nonreduced after extending the ground field. This does not happen for geometrically reduced algebraic spaces.
Definition 72.11.1. Let $k$ be a field. Let $X$ be an algebraic space over $k$.
Let $x \in |X|$ be a point. We say $X$ is geometrically reduced at $x$ if $\mathcal{O}_{X, \overline{x}}$ is geometrically reduced over $k$.
We say $X$ is geometrically reduced over $k$ if $X$ is geometrically reduced at every point of $X$.
Observe that if $X$ is geometrically reduced at $x$, then the local ring of $X$ at $x$ is reduced (Properties of Spaces, Lemma 66.22.6). Similarly, if $X$ is geometrically reduced over $k$, then $X$ is reduced (by Properties of Spaces, Lemma 66.21.4). The following lemma in particular implies this definition does not clash with the corresponding property for schemes over a field.
Lemma 72.11.2. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $x \in |X|$. The following are equivalent
$X$ is geometrically reduced at $x$,
for some étale neighbourhood $(U, u) \to (X, x)$ where $U$ is a scheme, $U$ is geometrically reduced at $u$,
for any étale neighbourhood $(U, u) \to (X, x)$ where $U$ is a scheme, $U$ is geometrically reduced at $u$.
Proof. Recall that the local ring $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$, see Properties of Spaces, Lemma 66.22.1. By Varieties, Lemma 33.6.2 we find that $U$ is geometrically reduced at $u$ if and only if $\mathcal{O}_{U, u}$ is geometrically reduced over $k$. Thus we have to show: if $A$ is a local $k$-algebra, then $A$ is geometrically reduced over $k$ if and only if $A^{sh}$ is geometrically reduced over $k$. We check this using the definition of geometrically reduced algebras (Algebra, Definition 10.43.1). Let $K/k$ be a field extension. Since $A \to A^{sh}$ is faithfully flat (More on Algebra, Lemma 15.45.1) we see that $A \otimes _ k K \to A^{sh} \otimes _ k K$ is faithfully flat (Algebra, Lemma 10.39.7). Hence if $A^{sh} \otimes _ k K$ is reduced, so is $A \otimes _ k K$ by Algebra, Lemma 10.164.2. Conversely, recall that $A^{sh}$ is a colimit of étale $A$-algebra, see Algebra, Lemma 10.155.2. Thus $A^{sh} \otimes _ k K$ is a filtered colimit of étale $A \otimes _ k K$-algebras. We conclude by Algebra, Lemma 10.163.7. $\square$
Lemma 72.11.3. Let $k$ be a field. Let $X$ be an algebraic space over $k$. The following are equivalent
$X$ is geometrically reduced,
for some surjective étale morphism $U \to X$ where $U$ is a scheme, $U$ is geometrically reduced,
for any étale morphism $U \to X$ where $U$ is a scheme, $U$ is geometrically reduced.
Proof. Immediate from the definitions and Lemma 72.11.2. $\square$
The notion isn't interesting in characteristic zero.
Lemma 72.11.4. Let $X$ be an algebraic space over a perfect field $k$ (for example $k$ has characteristic zero).
For $x \in |X|$, if $\mathcal{O}_{X, \overline{x}}$ is reduced, then $X$ is geometrically reduced at $x$.
If $X$ is reduced, then $X$ is geometrically reduced over $k$.
Proof. The first statement follows from Algebra, Lemma 10.43.6 and the definition of a perfect field (Algebra, Definition 10.45.1). The second statement follows from the first. $\square$
Lemma 72.11.5. Let $k$ be a field of characteristic $p > 0$. Let $X$ be an algebraic space over $k$. The following are equivalent
$X$ is geometrically reduced over $k$,
$X_{k'}$ is reduced for every field extension $k'/k$,
$X_{k'}$ is reduced for every finite purely inseparable field extension $k'/k$,
$X_{k^{1/p}}$ is reduced,
$X_{k^{perf}}$ is reduced, and
$X_{\bar k}$ is reduced.
Proof. Choose a surjective étale morphism $U \to X$ where $U$ is a scheme. Via Lemma 72.11.3 the lemma follows from the result for $U$ over $k$. See Varieties, Lemma 33.6.4. $\square$
Lemma 72.11.6. Let $k$ be a field. Let $X$ be an algebraic space over $k$. Let $k'/k$ be a field extension. Let $x \in |X|$ be a point and let $x' \in |X_{k'}|$ be a point lying over $x$. The following are equivalent
$X$ is geometrically reduced at $x$,
$X_{k'}$ is geometrically reduced at $x'$.
In particular, $X$ is geometrically reduced over $k$ if and only if $X_{k'}$ is geometrically reduced over $k'$.
Proof. Choose an étale morphism $U \to X$ where $U$ is a scheme and a point $u \in U$ mapping to $x \in |X|$. By Properties of Spaces, Lemma 66.4.3 we may choose a point $u' \in U_{k'} = U \times _ X X_{k'}$ mapping to both $u$ and $x'$. By Lemma 72.11.2 the lemma follows from the lemma for $U, u, u'$ which is Varieties, Lemma 33.6.6. $\square$
Lemma 72.11.7. Let $k$ be a field. Let $f : X \to Y$ be a morphism of algebraic spaces over $k$. Let $x \in |X|$ be a point with image $y \in |Y|$.
if $f$ is étale at $x$, then $X$ is geometrically reduced at $x$ $\Leftrightarrow $ $Y$ is geometrically reduced at $y$,
if $f$ is surjective étale, then $X$ is geometrically reduced $\Leftrightarrow $ $Y$ is geometrically reduced.
Proof. Part (1) is clear because $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{Y, \overline{y}}$ if $f$ is étale at $x$. Part (2) follows immediately from part (1). $\square$
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