# Thread: [SOLVED] Question in Ring theory....

1. ## [SOLVED] Question in Ring theory....

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.

2. Originally Posted by aurora
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 $R$ has order $p$ then it means $(R,+)$ (as a group) must be generated by an element, call it $a$, then it means $a,2a,3a,...,pa$ are all the elements of $R$. Notice that $(n\cdot a)(m\cdot a) = (m\cdot a)(n\cdot a)$ and so $R$ is a commutative ring. This is just te commutative part you still have other things to prove.

3. Right! I totally forgot about that.
I'll continue from here. If I still don't make it I'll know who to turn
Thank you!!!

4. Originally Posted by aurora

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.
we also need the ring to be unitary! (find a counter-example to your problem for rings without unity!) now suppose $R=\{0,a,2a, \cdots, (p-1)a \}.$ since $R$ is unitary, we have $a^2 \neq 0.$ (why?)

hence $R$ is an integral domain (why?) and we know that a finite integral domain is a field.

5. 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.

6. Originally Posted by aurora
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.
if you mean that your friend thinks that he has proved that any ring with p elements must be unitary, then i have to say your friend is flat wrong and ...

i don't know when finally people start believing in me!? anyway, here's a simple counter-example: $R=\left \{\begin{bmatrix}a & b \\ c & d \end{bmatrix} \in M_2(\mathbb{Z}_p): \ \ a=c=d=0 \right \}.$

7. Man, you're right!
hehe, my friend will hate me tomorow

Thank you!!!!

8. 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

9. Originally Posted by aurora
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
ok, in this case if $R=\{0,a,2a, \cdots, (p-1)a\},$ then since $R$ has no nonzero proper subgroup and we have that $R^2 \neq (0),$ we must have $R^2=R.$ thus $a=ka^2,$ for some $1 \leq k \leq p-1.$ hence

$ka=1_R.$ therefore $R$ is an integral domain (because $a^2 \neq 0$) and we know that a finite integral domain is a field. Q.E.D.

10. I thank you so much.