Consider
$$
\mathfrak q_3=(x^3,y^2-zw).
$$
It is $(x,y^2-zw)$-primary. A direct ideal-intersection calculation gives
$$
\begin{aligned}
\mathfrak q_1\cap\mathfrak q_2\cap\mathfrak q_3
&=x^2\big[(y,z)^2\cap(x,g)\big]\\
&=x^2\big[x(y,z)^2+g(y,z)\big]\\
&=(x^3y^2,x^3yz,x^3z^2,x^2y(y^2-zw),x^2z(y^2-zw))\\
&=I.
\end{aligned}
$$
This decomposition is irredundant and its radicals are
$$
(x),\qquad(y,z),\qquad(x,y^2-zw).
$$
Therefore
$$
\boxed{\operatorname{Ass}(k[x,y,z,w]/I)
=\{(x),(y,z),(x,y^2-zw)\}.}
$$
In particular, $\mathfrak p_3=(x,y^2-zw)$ is the embedded associated prime and the decomposition also confirms $\sqrt I=(x)\cap(y,z)$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 101 / 5 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-101/5/b/solution