Write the <affine scheme> as $U=\operatorname{Spec}B$. 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> $\mathfrak pB_{\mathfrak p}$. Hence
$$
f_x\notin\mathfrak m_x
\iff \bar f/1\notin\mathfrak pB_{\mathfrak p}
\iff \bar f\notin\mathfrak p.
$$
The last condition defines the <principal open subscheme> $D(\bar f)$, so
$$
\boxed{U\cap X_f=D(\bar f).}
$$
Every point of the <scheme> $X$ has such an affine neighbourhood. Thus $X_f$ is locally open, and hence is open in the <Zariski topology>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-113/1/a/solution