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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 $d = \deg (g_1)$, 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:

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

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