Let R be a commutative ring without unity (so |R|>1), P ideal in R. Can it be that R/P has unity?

I now see that if P=R then R/P has a unity. I observe that, in this case, P is a prime ideal. So let us repeat the question when P \noteq R and P prime. With this assumption R/P has no nonzero divisors of zero.

To look for an answer, I'll assume u+P is a unity in R/P, for some u in R. Then for all a, (u+P)(a+P) = a+P. u does not belong to P or else u+P=0 (P \noteq R implies |R/P|>1 impliies 0 \noteq 1 in R/P implies 0 \noteq u+P). I get ua-a belong to P. Fine if a \(\displaystyle \in\) P. If a \(\displaystyle \notin \) P, from the fact that R comm and P prime I obtain ua \(\displaystyle \notin \) P. I have gained nothing.

Another aproach perhaps comes from f: R --> R/P, f(r) = r+P. f is onto. And I know a surjective ring homomorphism from R to R', if both R and R' have unities 1 and 1', then the homom sends 1 into 1'. And here I'm stuck.

Any hint will be welcome. Good bye.