Let $R$ be a ring without a unit element and $R'$ be a (non trivial) ring with a unit element. Can there be an onto homomorphism from $R$ to $R'$?
Some observations: There cannot be an isomorphism, because then the element of $R$ which maps to $1$ in $R'$ must be a unit element in $R$. Also, we cannot have an onto homomorphism from a ring with unity to a ring without unity as $f(1)$ is going to be a unit element.
Edit: I am defining a ring homomorphism as a function $ \phi: R \rightarrow R'$ such that for all $a,b$ in $R$,
$$\phi(a+b) = \phi(a) + \phi(b) $$
$$\phi(ab) = \phi(a)\phi(b)$$
Best Answer
I'm surprised nobody suggested this obvious construction yet:
Let $R_1$ be a ring without identity, and $R_2$ be a ring with identity. Then $R=R_1\times R_2$ is a ring without identity, and $(r_1,r_2)\mapsto r_2$ is a surjective ring homomorphism of $R\to R_2$.