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 10.18.6, 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
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