## 29.10 Radicial and universally injective morphisms

In this section we define what it means for a morphism of schemes to be radicial and what it means for a morphism of schemes to be universally injective. We then show that these notions agree. The reason for introducing both is that in the case of algebraic spaces there are corresponding notions which may not always agree.

Definition 29.10.1. Let $f : X \to S$ be a morphism.

1. We say that $f$ is universally injective if and only if for any morphism of schemes $S' \to S$ the base change $f' : X_{S'} \to S'$ is injective (on underlying topological spaces).

2. We say $f$ is radicial if $f$ is injective as a map of topological spaces, and for every $x \in X$ the field extension $\kappa (x) \supset \kappa (f(x))$ is purely inseparable.

Lemma 29.10.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

1. For every field $K$ the induced map $\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), X) \to \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), S)$ is injective.

2. The morphism $f$ is universally injective.

3. The morphism $f$ is radicial.

4. The diagonal morphism $\Delta _{X/S} : X \longrightarrow X \times _ S X$ is surjective.

Proof. Let $K$ be a field, and let $s : \mathop{\mathrm{Spec}}(K) \to S$ be a morphism. Giving a morphism $x : \mathop{\mathrm{Spec}}(K) \to X$ such that $f \circ x = s$ is the same as giving a section of the projection $X_ K = \mathop{\mathrm{Spec}}(K) \times _ S X \to \mathop{\mathrm{Spec}}(K)$, which in turn is the same as giving a point $x \in X_ K$ whose residue field is $K$. Hence we see that (2) implies (1).

Conversely, suppose that (1) holds. Assume that $x, x' \in X_{S'}$ map to the same point $s' \in S'$. Choose a commutative diagram

$\xymatrix{ K & \kappa (x) \ar[l] \\ \kappa (x') \ar[u] & \kappa (s') \ar[l] \ar[u] }$

of fields. By Schemes, Lemma 26.13.3 we get two morphisms $a, a' : \mathop{\mathrm{Spec}}(K) \to X_{S'}$. One corresponding to the point $x$ and the embedding $\kappa (x) \subset K$ and the other corresponding to the point $x'$ and the embedding $\kappa (x') \subset K$. Also we have $f' \circ a = f' \circ a'$. Condition (1) now implies that the compositions of $a$ and $a'$ with $X_{S'} \to X$ are equal. Since $X_{S'}$ is the fibre product of $S'$ and $X$ over $S$ we see that $a = a'$. Hence $x = x'$. Thus (1) implies (2).

If there are two different points $x, x' \in X$ mapping to the same point of $s$ then (2) is violated. If for some $s = f(x)$, $x \in X$ the field extension $\kappa (s) \subset \kappa (x)$ is not purely inseparable, then we may find a field extension $\kappa (s) \subset K$ such that $\kappa (x)$ has two $\kappa (s)$-homomorphisms into $K$. By Schemes, Lemma 26.13.3 this implies that the map $\mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), X) \to \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), S)$ is not injective, and hence (1) is violated. Thus we see that the equivalent conditions (1) and (2) imply $f$ is radicial, i.e., they imply (3).

Assume (3). By Schemes, Lemma 26.13.3 a morphism $\mathop{\mathrm{Spec}}(K) \to X$ is given by a pair $(x, \kappa (x) \to K)$. Property (3) says exactly that associating to the pair $(x, \kappa (x) \to K)$ the pair $(s, \kappa (s) \to \kappa (x) \to K)$ is injective. In other words (1) holds. At this point we know that (1), (2) and (3) are all equivalent.

Finally, we prove the equivalence of (4) with (1), (2) and (3). A point of $X \times _ S X$ is given by a quadruple $(x_1, x_2, s, \mathfrak p)$, where $x_1, x_2 \in X$, $f(x_1) = f(x_2) = s$ and $\mathfrak p \subset \kappa (x_1) \otimes _{\kappa (s)} \kappa (x_2)$ is a prime ideal, see Schemes, Lemma 26.17.5. If $f$ is universally injective, then by taking $S'=X$ in the definition of universally injective, $\Delta _{X/S}$ must be surjective since it is a section of the injective morphism $X \times _ S X \longrightarrow X$. Conversely, if $\Delta _{X/S}$ is surjective, then always $x_1 = x_2 = x$ and there is exactly one such prime ideal $\mathfrak p$, which means that $\kappa (s) \subset \kappa (x)$ is purely inseparable. Hence $f$ is radicial. Alternatively, if $\Delta _{X/S}$ is surjective, then for any $S' \to S$ the base change $\Delta _{X_{S'}/S'}$ is surjective which implies that $f$ is universally injective. This finishes the proof of the lemma. $\square$

Proof. Combine Lemma 29.10.2 with the remark that $X \to S$ is separated if and only if the image of $\Delta _{X/S}$ is closed in $X \times _ S X$, see Schemes, Definition 26.21.3 and the discussion following it. $\square$

Lemma 29.10.4. A base change of a universally injective morphism is universally injective.

Proof. This is formal. $\square$

Lemma 29.10.5. A composition of radicial morphisms is radicial, and so the same holds for the equivalent condition of being universally injective.

Proof. Omitted. $\square$

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