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Let $A \subset B$ be an inclusion of commutative rings inducing $f: \text{Spec}(B) \rightarrow \text{Spec}(A)$. Must it be the case that $\text{Ass}(A) \subset f(\text{Ass}(B))$?

If this isn't true in general, is it true when $A,B$ are Noetherian rings? How generally can we do it?

For example, if $B$ is integral over $A$, then $\text{Spec}(B) \rightarrow \text{Spec}(A)$ is surjective. If $\mathfrak{p}$ is an associated prime of $A$, then write $\mathfrak{p} = \text{ann}(s)$. This is the image of some prime $\mathfrak{q} \subset B$, and an associated prime obtained from extending $\text{ann}(s)$ in $B_{\mathfrak{q}}$ should produce an associated prime which maps to $\mathfrak{q}$.

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    See [here](http://mathoverflow.net/questions/135783/lying-over-theorem-for-associated-primes).2013-07-24

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Let $A,B$ be noetherian rings and $\mathfrak p\in\operatorname{Ass}(A)$. We are looking for $P\in\operatorname{Ass}(B)$ such that $P\cap A=\mathfrak p$. Since $A\subseteq B$ we have $A_{\mathfrak p}\subseteq B_{\mathfrak p}$. Furthermore $\mathfrak p\in\operatorname{Ass}(A)$ implies $\mathfrak pA_{\mathfrak p}\in\operatorname{Ass}(A_{\mathfrak p})$. Thus we may assume that $A$ is local with maximal ideal $\mathfrak m\in\operatorname{Ass}(A)$. If $\mathfrak m=\operatorname{Ann}_A(a)$, then $\operatorname{Ann}_B(a)\neq B$ and let $P\in\operatorname{Spec}(B)$ minimal over $\operatorname{Ann}_B(a)$. It follows that $P\in\operatorname{Ass}(B)$ and $\operatorname{Ann}_A(a)\subseteq P\cap A$, so $P\cap A=\mathfrak m$.