Rings without unit.

Feb 2009
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0
Hi:
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.
 

Drexel28

MHF Hall of Honor
Nov 2009
4,563
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Berkeley, California
Hi:
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.
What about taking \(\displaystyle P\) to be a maximal ideal? Then, \(\displaystyle R/P\) is a field.
 
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Feb 2009
189
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What about taking \(\displaystyle P\) to be a maximal ideal? Then, \(\displaystyle R/P\) is a field.
This is very good news to me (that I can answer the question in the affirmative) and that important result did not come to my mind. Now I can go on working in the original problem, which was: Show that in a commutative ring R, an ideal P \noteq R is a prime ideal if and only if R/P is an intefral domain.

Maybe the existence of a nontrivial prime ideal guaranties that of a maximal ideal. Thanks a lot.
 

NonCommAlg

MHF Hall of Honor
May 2008
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an example is \(\displaystyle R=2\mathbb{Z}\) and \(\displaystyle I=6\mathbb{Z}.\) the identity element of \(\displaystyle R/I\) is \(\displaystyle 4 + I.\)

What about taking \(\displaystyle P\) to be a maximal ideal? Then, \(\displaystyle R/P\) is a field.
this is not necessarily true if \(\displaystyle R\) doesn't have identity element. for example in \(\displaystyle R=2\mathbb{Z}\) the ideal \(\displaystyle J=4\mathbb{Z}\) is maximal but \(\displaystyle R/J\) is not a field.
 
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