Suppose that I have two rings with identity, $S,R \in $ Ring, and a surjective $\mathbb Z$-algebra homomorphism $f: S \to R$ (hence $R$ is a quotient $\mathbb Z$-algebra of $S$).
Is it true that $R$ is also a quotient ring of $S$? I.e. can I choose a surjective $g: S \to R$ that is a ring homomorphism (hence it keeps the identity element)?
Note that if $f:S\to R$ is a surjective and multiplicative map between two rings $S$ and $R$ with identity, then $f(1_S)=1_R$. Indeed, if $s$ is a preimage of $1$, then $$ 1=f(s)=f(s \cdot 1)=f(s)f(1)=f(1). $$ In particular, a surjective, additive and multiplicative map between rings with identity is always a ring homomorphism.