The Stacks project

Lemma 26.6.1. Let $X$ be a locally ringed space. Let $Y$ be an affine scheme. Let $f \in \mathop{\mathrm{Mor}}\nolimits (X, Y)$ be a morphism of locally ringed spaces. Given a point $x \in X$ consider the ring maps

\[ \Gamma (Y, \mathcal{O}_ Y) \xrightarrow {f^\sharp } \Gamma (X, \mathcal{O}_ X) \to \mathcal{O}_{X, x} \]

Let $\mathfrak p \subset \Gamma (Y, \mathcal{O}_ Y)$ denote the inverse image of $\mathfrak m_ x$. Let $y \in Y$ be the corresponding point. Then $f(x) = y$.

Proof. Consider the commutative diagram

\[ \xymatrix{ \Gamma (X, \mathcal{O}_ X) \ar[r] & \mathcal{O}_{X, x} \\ \Gamma (Y, \mathcal{O}_ Y) \ar[r] \ar[u] & \mathcal{O}_{Y, f(x)} \ar[u] } \]

(see the discussion of $f$-maps below Sheaves, Definition 6.21.7). Since the right vertical arrow is local we see that $\mathfrak m_{f(x)}$ is the inverse image of $\mathfrak m_ x$. The result follows. $\square$

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Your email address will not be published.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 01HY. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 26.6.1 as pdf