Lemma 10.29.7. Let $R$ be a ring. The map $\mathop{\mathrm{Spec}}(R[x]) \to \mathop{\mathrm{Spec}}(R)$ is open, and the image of any standard open is a quasi-compact open.

Proof. It suffices to show that the image of a standard open $D(f)$, $f\in R[x]$ is quasi-compact open. The image of $D(f)$ is the image of $\mathop{\mathrm{Spec}}(R[x]_ f) \to \mathop{\mathrm{Spec}}(R)$. Let $\mathfrak p \subset R$ be a prime ideal. Let $\overline{f}$ be the image of $f$ in $\kappa (\mathfrak p)[x]$. Recall, see Lemma 10.17.9, that $\mathfrak p$ is in the image if and only if $R[x]_ f \otimes _ R \kappa (\mathfrak p) = \kappa (\mathfrak p)[x]_{\overline{f}}$ is not the zero ring. This is exactly the condition that $f$ does not map to zero in $\kappa (\mathfrak p)[x]$, in other words, that some coefficient of $f$ is not in $\mathfrak p$. Hence we see: if $f = a_ d x^ d + \ldots + a_0$, then the image of $D(f)$ is $D(a_ d) \cup \ldots \cup D(a_0)$. $\square$

Comment #5476 by damiano on

Just a typo: in the last sentence of the proof, there is a plus symbol missing after the \ldots: the formula $f = a_dx^d + \ldots a_0$, should be $f = a_dx^d + \ldots + a_0$!

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