Lemma 10.125.8. Let $R \to S$ be a ring homomorphism of finite presentation. Let $n \geq 0$. The set
is a quasi-compact open subset of $\mathop{\mathrm{Spec}}(S)$.
Lemma 10.125.8. Let $R \to S$ be a ring homomorphism of finite presentation. Let $n \geq 0$. The set
is a quasi-compact open subset of $\mathop{\mathrm{Spec}}(S)$.
Proof. It is open by Lemma 10.125.6. Let $S = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$ be a presentation of $S$. Let $R_0$ be the $\mathbf{Z}$-subalgebra of $R$ generated by the coefficients of the polynomials $f_ i$. Let $S_0 = R_0[x_1, \ldots , x_ n]/(f_1, \ldots , f_ m)$. Then $S = R \otimes _{R_0} S_0$. By Lemma 10.125.7 $V_ n$ is the inverse image of an open $V_{0, n}$ under the quasi-compact continuous map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(S_0)$. Since $S_0$ is Noetherian we see that $V_{0, n}$ is quasi-compact. $\square$
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