Hello, thanks for reading!
I'm having trouble with this next question, I'd really appreciate hints
Given that p is a prime, prove that any non-trivial ring (meaning, that it is not {0} ), with exactly p elements is a field.
I cannot seem to prove anything here... I'm trying to prove it's commutative, and cannot see how... let alone the other stuff...
Thank you.
Tomer.
If R had been unitary I would've solved it by now (thanks to Hacker).
I know how to prove R has no zero divisors if R is unitary...
But I'm positive it's not a given, and cannot find a counter-example as you suggested.
Mind you giving me that counter example?
Funny, a friend of mine told me yesterday he has proved R must have a unit but didn't mention how (I didn't work on it yesteday and didn't want him to "spoil" all the fun... ). Now he's asleep so I cannot contact him anyhow :-\
and THANK YOU.
oops... I think I was glad a little too early.
I failed to mention a given "detail" which I now seeas relevant .
We're actually not given only that R isn't {0}, but what's written is: R is non-trivial, meaning that not all products give "0".
I thought that's just the same as R not being {0} but I was obviously wrong. I'm sorry, I need to learn to write exactly what I'm given :-\
So... back on the problem again...
Sorry