Take
$$
f=T_1T_2-1.
$$
Then
$$
A=\mathbb R[T_1,T_2]/(T_1T_2-1)
\cong\mathbb R[t_1,t_1^{-1}].
$$
Every prime ideal in the image of
$$
\operatorname{Spec}A\longrightarrow\operatorname{Spec}\mathbb R[t_1]
$$
is disjoint from the multiplicative set $\{1,t_1,t_1^2,\ldots\}$. In particular, the prime ideal $(t_1)$ is not in the image, so the contraction map is not surjective.


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 4 / ii / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-101/4/ii/a/solution