A question related to associated prime ideals

Question

Let $f:A\to B$ be a (commutative) ring homomorphism, $f^*:\operatorname{Spec}A\leftarrow\operatorname{Spec}B$ the induced map, and $N$ a $B$-module. It is well known that $f^*(\operatorname{Ass}_BN)\subset\operatorname{Ass}_AN$, and that a strict inclusion is possible. It is also easy to show that $f^*(\operatorname{WeakAss}_BN)\supset\operatorname{WeakAss}_AN$, where $\operatorname{WeakAss}N$ is the set of prime ideals minimal over some $\operatorname{ann}n$, where $n\in N$. (ref: http://stacks.math.columbia.edu/tag/0546), and that again a strict inclusion is possible. Now let us define $\operatorname{ModerateAss}N$ to be the set of prime ideals of the form $\sqrt{\operatorname{ann} n}$, where $n\in N$. It is again straight forward to show that $f^*(\operatorname{ModerateAss}_BN)\subset\operatorname{ModerateAss}_AN$, but I couldn't think of an example of a strict inclusion. It is known that the inclusion is equality if $B$ or $N$ is Noetherian (ref: same as above), so the example should have non-Noetherian $B$ and $N$. So my question is: is strict inclusion $$f^*(\operatorname{ModerateAss}_BN)\varsubsetneq\operatorname{ModerateAss}_AN$$ possible ?

Answer 1

Yes; consider the injective ring homomorphism $$f:A=\mathbb{C}[x_1,x_2,x_3,\ldots]\to\mathbb{C}[x_1,y_1,x_2,y_2,\ldots]/(x_1y_1,x_2y_2,\ldots)=B.$$ It is easy to see that for any $b\in B$, $\sqrt{\operatorname{ann}_Bb}=\operatorname{ann}_Bb\subset B$ is not a prime ideal, so $\operatorname{ModerateAss}_BB=\varnothing$, hence $f^*(\operatorname{ModerateAss}_BB)=\varnothing$. On the other hand, for $\overline{y_1}\in B$, $$\sqrt{\operatorname{ann}_A\overline{y_1}}=\operatorname{ann}_A\overline{y_1}=(x_1)\subset A,$$ so $(x_1)\in\operatorname{ModerateAss}_AB$. Hence, $f^*(\operatorname{ModerateAss}_BB)\varsubsetneq\operatorname{ModerateAss}_AB$.