Subrings of ring homomorphisms Let $\phi: R \to S$ be a ring homomorphism, and suppose that $R$ and $S$ are rings with identity. 
Would $S' = \{r \in R\quad|\quad \phi(r) = 1_s\}$ be a subring of R?
More generally, how do subrings work for ring homomorphisms?
 A: A subring should contain the zero element; does $\{r\in R\mid \phi(r)=1_S\}$ contain the zero element of $R$? We should also have $\phi(0_R)=0_S$ by definition of a ring homomorphism, so we would then have to have $1_S=0_S$. There is only one ring $S$ which has this property, the "zero ring".
A: 
More generally, how do subrings work for ring homomorphisms?

The image of a ring homomorphism $\phi\colon R\to S$ is a subring of $S$. However, here you're asking if the preimage $\phi^{-1}(1_S)$ is a subring of $R$. As noted in other answers, unless $S$ is the zero ring, the set $\phi^{-1}(1_s)$ doesn't contain $0_R$, so it's not a subring of $R$. And unless $\phi$ is injective, it's not even a well defined function.
The structures that are preserved by preimages of homomorphisms are (left/right/twosided) ideals. Hence if $I$ is an ideal of $S$, then $\phi^{-1}(S)$ is an ideal of $R$.
But be careful, the image of an ideal under a ring homomorphism need not be an ideal. However, if $\phi$ is a surjective homomorphism, then $\phi(I)$ will be an ideal of $S$ whenever $I$ is an ideal of $R$.
