**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.
$\square$

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