TITLE: Can there be an onto homomorphism from a ring without unity to a ring with unity?
QUESTION [18 upvotes]: 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 $R$,
$$\phi(a+b) = \phi(a) + \phi(b) $$
$$\phi(ab) = \phi(a)\phi(b)$$

REPLY [2 votes]: For every $S \subseteq R$ denote by $\langle S \rangle$ the two-sided ideal generated by $S$, i.e. the smallest two-sided ideal containing $S$.
Then for every $x \in R$ one can consider the quotient ring
$$
  R'_x = R/\langle rx-r, xr-r \mid r \in R \rangle \,,
$$
for which $\overline{x} \in R'_x$ is a unit element.
Note that every unital ring $R'$ for which there exists a surjective ring homomorphism $\phi \colon R \to R'$ is a quotient of such a ring $R'_x$ (e.g. for $x \in R$ with $\phi(x) = 1_{R'}$).
Also note that for $x = 0$ we get the example $R'_x = 0$, which hasn’t been mentioned so far.