# If $B$ is an $A$-algebra of Noetherian rings then $\text{Ass}_{A}M=\{f^{-1}(p):p \in \text{Ass}_{B}M\}$ for a f.g. $B$-module $M.$

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.

• – Eric Wofsey Mar 14 at 4:19
• It really helps. Many thanks. – user371231 Mar 14 at 17:37