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I am having confusion regarding the relationship between the prime ideals of a ring and a certain quotient of a ring.

Let $A$ be a commutative ring with identity, and $a$ an ideal of $A$. Clearing there is a one-to-one order preserving correspondence between the ideals of $A/a$ and the ideals of $A$ containing $a$, but how does this relate to prime ideals in the respective rings? If $p$ is a prime ideal in $A$, under what conditions is its image in $A/a$ also prime?

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If $p$ contains $a$, then $p/a$ should be prime in $A/a$, since $(A/a)/(p/a)\cong A/p$, and so both are integral domains.

If $p$ does not contain $a$, then this need not hold: for example, let $A=\mathbb Z$, let $a=(2)$, $p=(3)$. Then $A/a=\mathbb Z/2\mathbb Z$, and the image of $(3)$ is the entirety of the quotient ring, which is not prime.

This counterexample is not ideal (no pun intended) since only a triviality prevents the ideal from being prime. If someone can think of a counterexample where the image is a nonprime proper ideal, that would be best.

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