The Stacks project

Proposition 41.6.3. Let $S$ is be a scheme. Let $\pi : X \to S$ be unramified and separated. Let $Y$ be an $S$-scheme and $y \in Y$ a point. Let $f, g : Y \to X$ be two $S$-morphisms. Assume

  1. $Y$ is connected

  2. $x = f(y) = g(y)$, and

  3. the induced maps $f^\sharp , g^\sharp : \kappa (x) \to \kappa (y)$ on residue fields are equal.

Then $f = g$.

Proof. The maps $f, g : Y \to X$ define maps $f', g' : Y \to X_ Y = Y \times _ S X$ which are sections of the structure map $X_ Y \to Y$. Note that $f = g$ if and only if $f' = g'$. The structure map $X_ Y \to Y$ is the base change of $\pi $ and hence unramified and separated also (see Morphisms, Lemmas 29.35.5 and Schemes, Lemma 26.21.12). Thus according to Theorem 41.6.2 it suffices to prove that $f'$ and $g'$ pass through the same point of $X_ Y$. And this is exactly what the hypotheses (2) and (3) guarantee, namely $f'(y) = g'(y) \in X_ Y$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 024V. Beware of the difference between the letter 'O' and the digit '0'.