Let $A,B$ Noetherian rings and $f: A \to B$ be a ring homomorphism. Let $M$ be a finitely generated $B$ module. Then want to show that $\text{Ass}_{A}M=\{f^{-1}(p):p \in \text{Ass}_{B}M\}.$

So far I could prove that RHS is contained in the LHS. If $p \in \text{Ass}_{B}M$ then $p=(0:_{B}x)$ for some $x \neq 0$ in $M.$ Then clearly $f^{-1}(p)=(0:_{A}x).$ Now for the converse let $q=(0:_{A}z) \in \text{Ass}_{A}M$. Then clearly $f^{-1}(0:_{B}z)=q,$ but failed to show that $(0:_{B}z) \in \text{Spec(B)}.$ I need some help. Thanks.


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