How to show this subset is proper? Let $f:R\to S$ a commutative ring homomorphism. I'm trying to prove that $Q^c=f^{-1}(Q)$ is a primary ideal if $Q$ is a primary ideal.
Curiously, I'm stuck only in the easiest part, to show that $Q^c$ is proper.
Any help is welcome.
thanks in advance.
 A: I assume that you take your rings to be unital, i.e. that they have multiplicative identities. If that is indeed the case, then recall that a homomorphism of unital rings is required to preserve these multiplicative identities, i.e. $f(1_R)=1_S$.
Let $f:R\to S$ be any homomorphism of unital rings, and let $I\subseteq S$ be any ideal. If $f^{-1}(I)=R$, then $1_R\in f^{-1}(I)$, so that $f(1_R)=1_S\in I$, so that $I=S$. Therefore, the preimage of a proper ideal is a proper ideal.
A: If you have a ring homomorphism $f:R\to S$ and an ideal $I\subseteq S$, then you have an induced injective ring homomorphism $\bar{f}: R/f^{-1}(I)\to S/I$. (If this isn't clear, then try to prove it!) Notice this also proves that $f^{-1}(I)$ is a proper ideal of $R$ if you're assuming that your rings are unital and that the map $f:R\to S$ is a unital ring homomorphism. (Explanation: If you've got a ring homomorphism $\bar{f}:R/f^{-1}(I)\to S/I$, then $R/f^{-1}(I)$ must be a ring with unity so it must be non-trivial, i.e., $f^{-1}(I)$ is a proper ideal in $R$.)
If $I\subseteq S$ is a primary ideal, then that's equivalent to stating that every zero divisor in $S/I$ is nilpotent. If you have an injective ring homomorphism $\bar{f}:R/f^{-1}(I)\to S/I$, then ...
The induced homomorphism $\bar{f}:R/f^{-1}(I)\to S/I$ gives rise to a factorization of the composition $R\to S\to S/I$ (where the first map is $f:R\to S$) as the composition $R\to R/f^{-1}(I)\to S/I$ of a surjective map followed by an injective map. It's always useful to have such factorizations in mind when proving theorems because it allows you to reduce your theorem to simpler and simpler cases.
