$ R \cong R s $ as rings if and only if $ s $ is invertible Let $ R $ be an integral domain and let $ s \in R $ be nonzero. Then why is it true that  $ R \cong R s $ if and only if $ s $ is invertible?
I can see that it is true for all nonzero $s$ if we consider $R$ and $Rs$ as $R$-modules using the bijection $r \mapsto rs$ but I can't see where the extra condition comes from in the ring case.
 A: *

*$\implies$: Let $\varphi : R \overset \sim \longrightarrow Rs$. Let $\varphi(1) = rs$. Then, $rs = 1$.

*$\impliedby$: Let $r_0 s = 1$. Then, let $\varphi:R \to Rs$ be the function that takes $r$ to $rr_0 s$. Then, $\varphi(rr') = rr' r_0 s = rr' = (r r_0 s) (r' r_0 s) = \varphi(r) \varphi(r')$, so $\varphi$ is a homomorphism. Now, let $\psi : Rs \to R$ be the function that sends $rs$ to $rs$. It is clear that $\varphi$ and $\psi$ are inverses of each other, so $R \cong Rs$.
A: I have some misgivings about the original statement of the question, but let me explain them as we go.
If $\phi$ is a ring isomorphism of $R\to Rs$, then it is nonzero, so it has to send $1$ to a nonzero idempotent of $Rs\subseteq R$. The only such element is $1$, so $\phi(1)=1\in Rs$. But this means, in particular, that $rs=1$ for some $r$, so that $s$ is a unit.
In the other direction, if $s$ is a unit, $R=Rs^{-1}s\subseteq Rs\subseteq R$, so that you have in fact $Rs=R$. Therefore the identity map is an isomorphism between $R$ and $s$. 

Now, I have a feeling the question was probably intended to be something more like this:

"Show that the map $r\mapsto rs$ is an isomorphism iff $s$ is a unit." 

Because we are working in a domain, it is automatically an injection for nonzero $s$. A further bifurcation could occur depending on whether the question was intended for ring homomorphisms or for module homomorphisms.

Now the problem is that $r\mapsto rs$ is only a ring homomorphism in one, trivial case. It would have to send identity to identity, so that $s$ is the identity of $Rs$, which is necessarily $1$, and idempotent. But a unit that is idempotent is $1$, so in fact $s=1$.

The question makes the most sense when interpreting $r\mapsto rs$ as a module homomorphism.  Then, indeed, for nonzero $s\in R$, we can say that $r\mapsto rs$ is a module isomorphism iff $s$ is a unit, and any unit $s\in R$ will do that, not just $1$.
The proof is straightforward: if $r\mapsto rs$ is to be onto, then something maps to $1$, therefore $rs=1$ for some $r$, and $s$ is a unit.
Conversely we know that if $s$ is a unit, left multiplication is $1-1$ already, and since $rss^{-1}$ maps to $rs$, we have that left multiplication is onto as well, hence a bijection. The fact that it is $R$-linear is elementary.
