The Stacks project

Theorem 10.29.10 (Chevalley's Theorem). Suppose that $R \to S$ is of finite presentation. The image of a constructible subset of $\mathop{\mathrm{Spec}}(S)$ in $\mathop{\mathrm{Spec}}(R)$ is constructible.

Proof. Write $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. We may factor $R \to S$ as $R \to R[x_1] \to R[x_1, x_2] \to \ldots \to R[x_1, \ldots , x_{n-1}] \to S$. Hence we may assume that $S = R[x]/(f_1, \ldots , f_ m)$. In this case we factor the map as $R \to R[x] \to S$, and by Lemma 10.29.6 we reduce to the case $S = R[x]$. By Lemma 10.29.1 suffices to show that if $T = (\bigcup _{i = 1\ldots n} D(f_ i)) \cap V(g_1, \ldots , g_ m)$ for $f_ i , g_ j \in R[x]$ then the image in $\mathop{\mathrm{Spec}}(R)$ is constructible. Since finite unions of constructible sets are constructible, it suffices to deal with the case $n = 1$, i.e., when $T = D(f) \cap V(g_1, \ldots , g_ m)$.

Note that if $c \in R$, then we have

\[ \mathop{\mathrm{Spec}}(R) = V(c) \amalg D(c) = \mathop{\mathrm{Spec}}(R/(c)) \amalg \mathop{\mathrm{Spec}}(R_ c), \]

and correspondingly $\mathop{\mathrm{Spec}}(R[x]) = V(c) \amalg D(c) = \mathop{\mathrm{Spec}}(R/(c)[x]) \amalg \mathop{\mathrm{Spec}}(R_ c[x])$. The intersection of $T = D(f) \cap V(g_1, \ldots , g_ m)$ with each part still has the same shape, with $f$, $g_ i$ replaced by their images in $R/(c)[x]$, respectively $R_ c[x]$. Note that the image of $T$ in $\mathop{\mathrm{Spec}}(R)$ is the union of the image of $T \cap V(c)$ and $T \cap D(c)$. Using Lemmas 10.29.5 and 10.29.6 it suffices to prove the images of both parts are constructible in $\mathop{\mathrm{Spec}}(R/(c))$, respectively $\mathop{\mathrm{Spec}}(R_ c)$.

Let us assume we have $T = D(f) \cap V(g_1, \ldots , g_ m)$ as above, with $\deg (g_1) \leq \deg (g_2) \leq \ldots \leq \deg (g_ m)$. We are going to use induction on $m$, and on the degrees of the $g_ i$. Let $d_1 = \deg (g_1)$, i.e., $g_1 = c x^{d_1} + l.o.t$ with $c \in R$ not zero. Cutting $R$ up into the pieces $R/(c)$ and $R_ c$ we either lower the degree of $g_1$ (and this is covered by induction) or we reduce to the case where $c$ is invertible. If $c$ is invertible, and $m > 1$, then write $g_2 = c' x^{d_2} + l.o.t$. In this case consider $g_2' = g_2 - (c'/c) x^{d_2 - d_1} g_1$. Since the ideals $(g_1, g_2, \ldots , g_ m)$ and $(g_1, g_2', g_3, \ldots , g_ m)$ are equal we see that $T = D(f) \cap V(g_1, g_2', g_3\ldots , g_ m)$. But here the degree of $g_2'$ is strictly less than the degree of $g_2$ and hence this case is covered by induction.

The bases case for the induction above are the cases (a) $T = D(f) \cap V(g)$ where the leading coefficient of $g$ is invertible, and (b) $T = D(f)$. These two cases are dealt with in Lemmas 10.29.9 and 10.29.7. $\square$


Comments (4)

Comment #4613 by Noah Olander on

I think this is a regular induction, not a descending induction.

Comment #5091 by Runyu on

Do we assume to be noetherian here? If we assume so, we do not need to say "finite presentation", if not, why would a locally closdled set look like ?

Comment #5300 by on

No we do not assume is Noetherian. Please read earlier in the section to find the answer to your question.

There are also:

  • 6 comment(s) on Section 10.29: Images of ring maps of finite presentation

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 00FE. Beware of the difference between the letter 'O' and the digit '0'.