Are ring homomorphisms always unital if the rings have $1_{R}$

Question

I know according to some textbooks, rings do not have to contain $1$. But if I define rings to have $1$, are all ring homomorphisms unital? Here's my attempt to prove this:

Let $\phi:R\rightarrow S$ be a ring homomorphism. $\forall x\in R$, $\phi(x)=\phi(1\cdot x)=\phi(1)\phi(x)\implies \phi(1)=1$.

Is there anything wrong with my proof? If my proof is correct, then what's the point of classifying unital and non-unital ring homomorphisms? There are a lot of good reasons why rings should include $1$. It is better to for rings to have $1$, so according to this definition, all ring homomorphisms are unital.

Answer 1

u could only deduce that $\phi(x) (1-\phi(1))=0$ for every $x\in R$(there is no cancellation for multplication in general) , if $S$ is an integral domain and $\phi$ isn't the zero map, , then u can conclude $\phi(1)=1$,so u need extra assumption on the ring , or u define ring homomorphisms to satisfy $\phi(1) =1$.

Answer 2

Here is a concrete example of what can go wrong.

Consider the ring $\mathbf Z/(6)$. Its subset $\{0, 3 \bmod 6\}$ is a "subset with ring structure" since $3+3 = 0$ and $(3)(3) = 3$, so $3$ is the multiplicative identity of that two-element ring. So $f \colon \mathbf Z/(2) \to \mathbf Z/(6)$ by $f(x \bmod 2) = 3x \bmod 6$ is an additive and multiplicative mapping (because $3^2 = 3$ in $\mathbf Z/(6)$). It does not send the multiplicative identity of $\mathbf Z/(2)$ to the multiplicative identity of $\mathbf Z/(6)$.