Sorry if this one was already asked, I did not find it. Here is the story:
Let us take a unitary ring $R$, the notion of unit element is very clear: $$r \in R^{\times} \iff \exists r' \in R \mid rr' = 1 = r'r$$ Of course one can lighten this statement with only left or right invertibility in the finite case (due to the finiteness of endomorphism rings).
For finite unitary rings we also have the result $\forall r \in R$, $r$ is either a unit or a divisor (left and right) of zero.
This can be achieved by studying the ring morphisms:
$$\begin{array}{rcl}
R & \to & \mathrm{End}_{\mathrm{Mod}-R} (R_R) \\
r & \mapsto & (\tau_r: s \mapsto rs)
\end{array}$$
Being a unit is then equivalent for $\tau_r$ to be an automorphism an a zero divisor
Now when we study these same ring morphism (that become rng morphisms) when $R$ is non-unitary, we also have the result if $R$ is finite: either $R$ is unitary or $R$ has only zero divisors.
This can be proven since we have an equivalence: $i\in R$ is a unit for $R$ $\iff$ $\tau_i$ is the identity. But I can prove it only when $R$ is finite.
My 2 (very) related questions:
Q1 (answered): **Are there examples of (infinite) non unitary rings which have a unit on the left but not on the right? **(i.e. $\tau_r = \mathrm{id}_{R_R}$ but $(\tau'_r :s \mapsto rs) \neq \mathrm{id}_{{}_R\! R}$).
And as @rschwieb stated if we have both left and right units, they are the same.
Thanks to @IzaakvanDongen, we have the answer, there are examples of such right but not left identity and we can have multiple left (or right) identities.
My main motivation is to find a somewhat "canonical" isomorphism from the following morphism: $$\begin{array}{rcl} \hom_{\mathrm{Mod}-R}(R_R ,X_R)& \to& X_R \\ f & \mapsto & f(i) \end{array}$$ Which would then be bijective.
Q2: Wether or not we have these left or right unit, if the have a rng $R$, if we have left or right invertible elements (with this following definition):
Def: $r\in R$ is quasi left (resp. right) invertible on the left if $\tau_r$ is a monomorphism (resp. epimorphism) and left (resp. right) invertible on the left if $\tau_r$ is a split monomorphism (resp. epimorphism).
Of course if there is a retraction (or section) of the form $\tau_{r'}$ split (or epi) this definition is equivalent to have $r'$ such that $r'r = i$ where $i$ is the unit on the left (so it has to exist). But since it is not necessarily the case, do we have example of such modules with left (quasi)-invertible elements on the left?
Some motivation for this definition:
For the previous "altered morphism" $f \in \hom(R_R,X_R) \mapsto f(r)\in X$ we would have $r$ quasi right invertible on the left $\implies$ this morphism is injective, since knowing $f(r)$ implies knowing $f\circ \tau_r$.
Furthermore if it is right invertible on the left ($\tau_r$ split epi with section $s \in \mathrm{End}(R_R)$) the association, $x \mapsto (t \in R \mapsto x \cdot s(t))$ is an right inverse map of our injective ring morphism.
I would like to say it is an iso but this problem looks a lot like studying the representable functors:
Here our "representables" would be the morphisms $x^\wedge : R_R \to X$ that send $r$ on $x \cdot r$, the problem is how can we retrieve $x$ from this? This situation is not likely to have a solution without a proper unit on the left.
Now if $r$ is left invertible on the left, $\tau_r$ has a retraction $\mu$ one should have surjectivity on our iso?
So don't hesitate to correct me if I'm wrong, but given a quasi-right inverse on the left $r$, we have an injective ring morphism: $$\begin{array}{rcl} \hom_{\mathrm{Mod}-R}(R_R ,X_R)& \to& X_R \\ f & \mapsto & f(r) \end{array}$$ Furthermore if this quasi-right inverse on the left is a right inverse on the left, this is an isomorphism of rings with inverse: $$\begin{array}{rcl} X_R& \to& \hom_{\mathrm{Mod}-R}(R_R ,X_R) \\ x & \mapsto & (t \mapsto x \cdot s(r)) \end{array}$$ where $s$ is a section of the split epimorphism $\tau_r$.
PS: These two questions have motivated this present question:
