Lemma 32.8.2. Notation and assumptions as in Situation 32.8.1. If $f$ is affine, then there exists an index $i \geq 0$ such that $f_ i$ is affine.

Proof. Let $Y_0 = \bigcup _{j = 1, \ldots , m} V_{j, 0}$ be a finite affine open covering. Set $U_{j, 0} = f_0^{-1}(V_{j, 0})$. For $i \geq 0$ we denote $V_{j, i}$ the inverse image of $V_{j, 0}$ in $Y_ i$ and $U_{j, i} = f_ i^{-1}(V_{j, i})$. Similarly we have $U_ j = f^{-1}(V_ j)$. Then $U_ j = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_{j, i}$ (see Lemma 32.2.2). Since $U_ j$ is affine by assumption we see that each $U_{j, i}$ is affine for $i$ large enough, see Lemma 32.4.13. As there are finitely many $j$ we can pick an $i$ which works for all $j$. Thus $f_ i$ is affine for $i$ large enough, see Morphisms, Lemma 29.11.3. $\square$

## Comments (0)

There are also:

• 5 comment(s) on Section 32.8: Descending properties of morphisms

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Your email address will not be published.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