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

Write the affine scheme as $U=\mathrm{Spec}B$. A point $x\in U$ corresponds to a prime ideal $\mathfrak{p}\subset B$, and its local ring is ${\mathcal{O}}_{X,x}\cong B_{\mathfrak{p}}$ with maximal ideal $\mathfrak{p}{B}_{\mathfrak{p}}$. HenceThe last condition defines the principal open subscheme $D(\bar{f})$, soEvery point of the scheme $X$ has such an affine neighbourhood. Thus $X_f$ is locally open, and hence is open in the Zariski topology.

Choose an open cover of $X$ by affine opens $U_i=\mathrm{Spec}{B}_i$. 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 meansso $f_i^{n_i}{a}_i=0$ for some $n_i\ge 0$. Taking $n>0$ at least as large as every $n_i$ gives $(f^{n}a){|}_{U_i}=0$ on every member of the finite cover. The local identity axiom for the sheaf of rings ${\mathcal{O}}_X$ therefore gives

Let $X={\bigcup }_{i=1}^r{U}_i$ be the stated finite affine cover. By the description of sections on a principal open subscheme, after increasing denominators separately we may writefor some $a_i\in \Gamma (U_i,{\mathcal{O}}_X)$. Multiplying the numerators by powers of $f$ lets us use one exponent $n$ for every $i$.

On $W_{ij}=U_i\cap U_j$, the section $a_i-a_j$ vanishes after restriction to $(W_{ij}{)}_f$. Each $W_{ij}$ is quasi-compact, so part (b) supplies $m_{ij}$ with $f^{m_{ij}}(a_i-a_j)=0$ on $W_{ij}$. Choose one $m$ valid for all the finitely many pairs. The sections $f^m{a}_i$ now agree on every overlap, and the gluing axiom of a sheaf of rings produces $a\in \Gamma (X,{\mathcal{O}}_X)$ with $a{|}_{U_i}=f^m{a}_i$. On $X_f$,Thus some power of $f$ times $b$ extends to a global regular function on $X$.

Restriction and division by powers of $f$ define a natural ring homomorphismIf $\Phi (a/f^n)=0$, then $a{|}_{X_f}=0$, and part (b) gives $f^{m}a=0$ for some $m$; this is exactly the criterion that $a/f^n=0$ in the localization of a ring $A_f$. Hence $\Phi$ is injective. Given $b$ in the target, part (c) gives $f^{N}b=a{|}_{X_f}$ for some $a\in A$, so $b=\Phi (a/f^N)$. Hence $\Phi$ is surjective and