Lemma 99.14.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a universally injective morphism of algebraic stacks. Let $y : \mathop{\mathrm{Spec}}(k) \to \mathcal{Y}$ be a morphism where $k$ is an algebraically closed field. If $y$ is in the image of $|\mathcal{X}| \to |\mathcal{Y}|$, then there is a morphism $x : \mathop{\mathrm{Spec}}(k) \to \mathcal{X}$ with $y = f \circ x$.

**Proof.**
We first remark this lemma is not a triviality, because the assumption that $y$ is in the image of $|f|$ means only that we can lift $y$ to a morphism into $\mathcal{X}$ after possibly replacing $k$ by an extension field. To prove the lemma we may base change $f$ by $y$, hence we may assume we have a nonempty algebraic stack $\mathcal{X}$ and a universally injective morphism $\mathcal{X} \to \mathop{\mathrm{Spec}}(k)$ and we want to find a $k$-valued point of $\mathcal{X}$. We may replace $\mathcal{X}$ by its reduction. We may choose a field $k'$ and a surjective, flat, locally finite type morphism $\mathop{\mathrm{Spec}}(k') \to \mathcal{X}$, see Properties of Stacks, Lemma 98.11.2. Since $\mathcal{X} \to \mathop{\mathrm{Spec}}(k)$ is universally injective, we find that

is surjective as the base change of the surjective morphism $\Delta : \mathcal{X} \to \mathcal{X} \times _{\mathop{\mathrm{Spec}}(k)} \mathcal{X}$ (Lemma 99.14.5). Since $k$ is algebraically closed $k' \otimes _ k k'$ is a domain (Algebra, Lemma 10.49.4). Let $\xi \in \mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k')$ be a point mapping to the generic point of $\mathop{\mathrm{Spec}}(k' \otimes _ k k')$. Let $U$ be the reduced induced closed subscheme structure on the connected component of $\mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k')$ containing $\xi $. Then the two projections $U \to \mathop{\mathrm{Spec}}(k')$ are locally of finite type, as this was true for the projections $\mathop{\mathrm{Spec}}(k') \times _\mathcal {X} \mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k')$ as base changes of the morphism $\mathop{\mathrm{Spec}}(k') \to \mathcal{X}$. Applying Varieties, Proposition 33.31.1 we find that the integral closures of the two images of $k'$ in $\Gamma (U, \mathcal{O}_ U)$ are equal. Looking in $\kappa (\xi )$ means that any element of the form $\lambda \otimes 1$ is algebraically dependend on the subfield

Since $k$ is algebraically closed, this is only possible if $k' = k$ and the proof is complete. $\square$

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