Ring homomorphisms inducing same map on their affine schemes Let $f,g:R \to S$ be ring homomorphisms between commutative rings $R,S$.
Suppose that  there are unit elements of $R$ and $S$ and we denote the units by $r \in R,s \in S$, i.e there are some $r' \in R,s' \in S$ such that $rr' =1,ss'=1$.
If$f(x) = s \times g(r \times x), x \in R$, then $f,g$ induce same map on $\operatorname{Spec} S \to \operatorname{Spec} R$. Is there other possibility that ring homomorphism inducing same map on their spectrums?
 A: In the context of algebraic geometry our rings are all unital rings and ring homomorphisms should send $1\mapsto 1$. So in the $f,g$ you have in your question you are getting $f(1)=s\cdot g(r\cdot 1)=s\cdot g(r)$ which need not equal $1$, so this $f$ isn't actually a ring homomorphism unless $s=g(r)^{-1}$ but in this case you are getting $f=g$ anyways.
Thus, I am going to interpret your question as follows:

Question: What are examples of different ring homomorphisms $R\to S$ inducing the same map $\operatorname{Spec}(S)\to\operatorname{Spec}(R)$?

Here are some typical examples that come to mind:

*

*If you take a field $F$ with a nontrivial automorphism $\sigma:F\to F$, then $\sigma$ and the identity $\mathrm{id}_F$ are different ring homomorphisms, but $\operatorname{Spec}(F)$ consists of a single point so they must induce the same map $\operatorname{Spec}(F)\to\operatorname{Spec}(F)$.


*If $R$ is a ring of characteristic $p$ then one has the Frobenius $\phi:R\to R$ sending $r\mapsto r^p$. You can check this induces the identity on $\operatorname{Spec}(R)$ so as long as $\phi\neq\mathrm{id}_R$ for your ring $R$ you have that these are distinct homomorphisms inducing the same map on $\operatorname{Spec}(R)$.
