Lemma 68.16.5. In Situation 68.16.1 the morphism $p : X \to \mathop{\mathrm{Spec}}(A)$ is universally injective.

**Proof.**
Let $A \to k$ be a ring homomorphism where $k$ is a field. It suffices to show that $\mathop{\mathrm{Spec}}(k) \times _{\mathop{\mathrm{Spec}}(A)} X$ has at most one point (see Morphisms of Spaces, Lemma 66.19.6). Using Lemma 68.16.3 we may assume that $A$ is a field and we have to show that $|X|$ has at most one point.

Let's think of $X$ as an algebraic space over $\mathop{\mathrm{Spec}}(k)$ and let's use the notation $X(K)$ to denote $K$-valued points of $X$ for any extension $K/k$, see Morphisms of Spaces, Section 66.24. If $K/k$ is an algebraically closed field extension of large transcendence degree, then we see that $X(K) \to |X|$ is surjective, see Morphisms of Spaces, Lemma 66.24.2. Hence, after replacing $k$ by $K$, we see that it suffices to prove that $X(k)$ is a singleton (in the case $A = k)$.

Let $x, x' \in X(k)$. By Decent Spaces, Lemma 67.14.4 we see that $x$ and $x'$ are closed points of $|X|$. Hence $x$ and $x'$ map to distinct points of $\mathop{\mathrm{Spec}}(k)$ if $x \not= x'$ by Lemma 68.16.4. We conclude that $x = x'$ as desired. $\square$

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