When is the preimage of prime ideal is not a prime ideal? If $f\colon R\to S$ is a ring homomorphism such that $f(1)=1$, it's straightforward to show that the preimage of a prime ideal is again a prime ideal.
What happens though if $f(1)\neq 1$? I use the fact that $f(1)=1$ to show that the preimage of a prime ideal is proper, so I assume there is some example where the preimage of a prime ideal is not proper, and thus not prime when $f(1)\neq 1$? Could someone enlighten me on such an example? 

First, I object to the fact that you consider maps which don't take $1_R$ to $1_S$ to be ring homomorphisms. 
That said, yes if you were to consider such maps there are many examples. For example let $R$ be any integral domain, $S = R\oplus R$ and $f:R \rightarrow S$ be the embedding of $R$ into the first coordinate in $S$. Then $f(R)$ is a prime ideal of $S.$ And voila a counterexample.  
