Definition 76.3.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. We say $f$ is radicial if for any morphism $\mathop{\mathrm{Spec}}(K) \to Y$ where $K$ is a field the reduction $(\mathop{\mathrm{Spec}}(K) \times _ Y X)_{red}$ is either empty or representable by the spectrum of a purely inseparable field extension of $K$.
76.3 Radicial morphisms
It turns out that a radicial morphism is not the same thing as a universally injective morphism, contrary to what happens with morphisms of schemes. In fact it is a bit stronger.
Lemma 76.3.2. A radicial morphism of algebraic spaces is universally injective.
Proof. Let $S$ be a scheme. Let $f : X \to Y$ be a radicial morphism of algebraic spaces over $S$. It is clear from the definition that given a morphism $\mathop{\mathrm{Spec}}(K) \to Y$ there is at most one lift of this morphism to a morphism into $X$. Hence we conclude that $f$ is universally injective by Morphisms of Spaces, Lemma 67.19.2. $\square$
Example 76.3.3. It is no longer true that universally injective is equivalent to radicial. For example the morphism of Spaces, Example 65.14.7 is universally injective, but is not radicial in the sense above.
\[ X = [\mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})/ \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})] \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{Q}) \]
Nonetheless it is often the case that the reverse implication holds.
Lemma 76.3.4. Let $S$ be a scheme. Let $f : X \to Y$ be a universally injective morphism of algebraic spaces over $S$.
If $f$ is decent then $f$ is radicial.
If $f$ is quasi-separated then $f$ is radicial.
If $f$ is locally separated then $f$ is radicial.
Proof. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is stable under base change and composition and holds for closed immersions. Assume $f : X \to Y$ has $\mathcal{P}$ and is universally injective. Then, in the situation of Definition 76.3.1 the morphism $(\mathop{\mathrm{Spec}}(K) \times _ Y X)_{red} \to \mathop{\mathrm{Spec}}(K)$ is universally injective and has $\mathcal{P}$. This reduces the problem of proving
\[ \mathcal{P} + \text{universally injective} \Rightarrow \text{radicial} \]
to the problem of proving that any nonempty reduced algebraic space $X$ over field whose structure morphism $X \to \mathop{\mathrm{Spec}}(K)$ is universally injective and $\mathcal{P}$ is representable by the spectrum of a field. Namely, then $X \to \mathop{\mathrm{Spec}}(K)$ will be a morphism of schemes and we conclude by the equivalence of radicial and universally injective for morphisms of schemes, see Morphisms, Lemma 29.10.2.
Let us prove (1). Assume $f$ is decent and universally injective. By Decent Spaces, Lemmas 68.17.4, 68.17.6, and 68.17.2 (to see that an immersion is decent) we see that the discussion in the first paragraph applies. Let $X$ be a nonempty decent reduced algebraic space universally injective over a field $K$. In particular we see that $|X|$ is a singleton. By Decent Spaces, Lemma 68.14.2 we conclude that $X \cong \mathop{\mathrm{Spec}}(L)$ for some extension $K \subset L$ as desired.
A quasi-separated morphism is decent, see Decent Spaces, Lemma 68.17.2. Hence (1) implies (2).
Let us prove (3). Recall that the separation axioms are stable under base change and composition and that closed immersions are separated, see Morphisms of Spaces, Lemmas 67.4.4, 67.4.8, and 67.10.7. Thus the discussion in the first paragraph of the proof applies. Let $X$ be a reduced algebraic space universally injective and locally separated over a field $K$. In particular $|X|$ is a singleton hence $X$ is quasi-compact, see Properties of Spaces, Lemma 66.5.2. We can find a surjective étale morphism $U \to X$ with $U$ affine, see Properties of Spaces, Lemma 66.6.3. Consider the morphism of schemes
\[ j : U \times _ X U \longrightarrow U \times _{\mathop{\mathrm{Spec}}(K)} U \]
As $X \to \mathop{\mathrm{Spec}}(K)$ is universally injective $j$ is surjective, and as $X \to \mathop{\mathrm{Spec}}(K)$ is locally separated $j$ is an immersion. A surjective immersion is a closed immersion, see Schemes, Lemma 26.10.4. Hence $R = U \times _ X U$ is affine as a closed subscheme of an affine scheme. In particular $R$ is quasi-compact. It follows that $X = U/R$ is quasi-separated, and the result follows from (2). $\square$
Remark 76.3.5. Let $X \to Y$ be a morphism of algebraic spaces. For some applications (of radicial morphisms) it is enough to require that for every $\mathop{\mathrm{Spec}}(K) \to Y$ where $K$ is a field
the space $|\mathop{\mathrm{Spec}}(K) \times _ Y X|$ is a singleton,
there exists a monomorphism $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K) \times _ Y X$, and
$K \subset L$ is purely inseparable.
If needed later we will may call such a morphism weakly radicial. For example if $X \to Y$ is a surjective weakly radicial morphism then $X(k) \to Y(k)$ is surjective for every algebraically closed field $k$. Note that the base change $X_{\overline{\mathbf{Q}}} \to \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$ of the morphism in Example 76.3.3 is weakly radicial, but not radicial. The analogue of Lemma 76.3.4 is that if $X \to Y$ has property ($\beta $) and is universally injective, then it is weakly radicial (proof omitted).
Lemma 76.3.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume
$f$ is locally of finite type,
for every étale morphism $V \to Y$ the map $|X \times _ Y V| \to |V|$ is injective.
Then $f$ is universally injective.
Proof. The question is étale local on $Y$ by Morphisms of Spaces, Lemma 67.19.6. Hence we may assume that $Y$ is a scheme. Then $Y$ is in particular decent and by Decent Spaces, Lemma 68.18.9 we see that $f$ is locally quasi-finite. Let $y \in Y$ be a point and let $X_ y$ be the scheme theoretic fibre. Assume $X_ y$ is not empty. By Spaces over Fields, Lemma 72.10.8 we see that $X_ y$ is a scheme which is locally quasi-finite over $\kappa (y)$. Since $|X_ y| \subset |X|$ is the fibre of $|X| \to |Y|$ over $y$ we see that $X_ y$ has a unique point $x$. The same is true for $X_ y \times _{\mathop{\mathrm{Spec}}(\kappa (y))} \mathop{\mathrm{Spec}}(k)$ for any finite separable extension $k/\kappa (y)$ because we can realize $k$ as the residue field at a point lying over $y$ in an étale scheme over $Y$, see More on Morphisms, Lemma 37.35.2. Thus $X_ y$ is geometrically connected, see Varieties, Lemma 33.7.11. This implies that the finite extension $\kappa (x)/\kappa (y)$ is purely inseparable.
We conclude (in the case that $Y$ is a scheme) that for every $y \in Y$ either the fibre $X_ y$ is empty, or $(X_ y)_{red} = \mathop{\mathrm{Spec}}(\kappa (x))$ with $\kappa (y) \subset \kappa (x)$ purely inseparable. Hence $f$ is radicial (some details omitted), whence universally injective by Lemma 76.3.2. $\square$
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