Choose an <open cover> of $X$ by affine opens $U_i=\operatorname{Spec}B_i$. Since $X$ is a <quasi-compact topological space>, finitely many suffice. Let $f_i,a_i\in B_i$ be the restrictions of $f,a$. Part (a) identifies $U_i\cap X_f$ with $D(f_i)$, whose ring of <regular functions> is the <localization of a ring> $(B_i)_{f_i}$. The vanishing of $a$ there means
$$
a_i/1=0\quad\hbox{in }(B_i)_{f_i},
$$
so $f_i^{n_i}a_i=0$ for some $n_i\geq0$. Taking $n>0$ at least as large as every $n_i$ gives $(f^na)|_{U_i}=0$ on every member of the finite cover. The local identity axiom for the <sheaf of rings> $\mathcal O_X$ therefore gives
$$
\boxed{f^na=0\text{ in }\Gamma(X,\mathcal O_X).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 113 / 1 / b / Solution

{wiki=Quasi-compact_space}

= Quasi-compact
{synonym}

A topological space is quasi-compact when every <open cover> has a finite subcover. This is the form of compactness used for spaces that need not be Hausdorff, especially spectra and schemes.

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 Incoming links: Quasi-compact topological space