Let $A$ be a commutative ring and $X = \text{Spec}(A)$. The closed sets are those of the form $V(E) = \{$ prime ideals $\hat{p} \subset A $ containing $E \}$. And the open sets are the complements of these.
I've already shown that $X$ is compact by showing that $1$ is in a certain ideal and so must equal a finite sum.
How do you show that in general $X_f = X \setminus V(f)$ is compact in the Zariski topology?
I tried generalizing: that it's true of any basis element in a compact topological space. But that never left the runway.
Hints please.
Best Answer
Your generalization is false, since for example open arcs in $S^1$ form a basis for its topology, but are not compact.
You could use there's a homeomorphism ${\rm Spec}(A_f)\simeq X_f$, and use what you've already proven.
You can also give a direct proof, in the same lines as the one for $X$. Suppose we cover $X_f$ with the basic open sets $X_{f_i}$. Since $X_f\cap X_{f_i}= X_{ff_i}$, we may suppose that $X_f=\bigcup_i X_{g_i}$, and $g_i\in (f)$. Translating into open sets, we get that $$V((g_i:i\in I))=V(f)$$
This means there is an equation of the form $f^n=a_1g_1+\cdots+a_ng_n$. This means that $f$ is in the radical of $(g_1,\ldots,g_n)$ so that $V(f)=V(g_1,\ldots,g_n)$ (since we already had $g_i\in (f)$), so that indeed $X_f=\bigcup\limits_{i=1}^n X_{g_i}$.