Question:
Suppose the quotient ring $R/I$ has a non-zero identity (where $I$ is a proper ideal of the ring $R$). Does this imply the ring $R$ has an identity too?
Context:
I'm aware the converse is true: that if $R$ has an identity $1_R\not=0$, then $R/I$ has the identity $1_R+I$ which is nonzero if $I\not=R$ (since if $1_R+I=0_R+I$ is zero, then $1_R\in I$ which would imply $I=R$).
This came up when I was working through the proof that the quotient of a commutative ring by a prime ideal is an integral domain. Clearly in this case, the quotient ring $R/P$ must have an identity since that's part of the definition of an integral domain. But it seemed like the result could hold true even if $R$ failed to have an identity.
My attempt:
So let's suppose $R/I$ has a non-zero identity $r+I$ for some $r\in R$. Being the identity means that $$(r+I)(s+I):=rs+I=s+I$$ for every $s\in R$. This implies that $rs-s\in I$ for every $s\in R$.
One observation is that since $r+I$ is non-zero, we must have $r\not\in I$. But if we pick, in particular, $s=r$ we deduce that $r^2-r\in I$. This implies that $r^2\not\in I$ as well (since otherwise, $r^2\in I$ and $r^2-r\in I$ together imply $r\in I$). We can similarly pick $s=r^2$ to show that $r^3-r^2\in I$ implying that $r^3\not\in I$ as well. And by induction, that $r^n\not\in I$ for all $n\in \mathbb{N}^+$. But I'm not sure where to go from here...
If it is possible for a identity-less ring to have an quotient with identity, what's the simplest example?
Consider $R=3\Bbb Z$, $I=6\Bbb Z$ then $R/I \cong \Bbb F_2$, since $3\cdot 3 =9 \sim 3$. But $R$ is nonunital.