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Math Help - [SOLVED] Question in Ring theory....

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

    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.
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  2. #2
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    Quote Originally Posted by aurora View Post
    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 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.
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    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!!!
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  4. #4
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    Quote Originally Posted by aurora View Post

    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.
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    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.
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  6. #6
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    Quote Originally Posted by aurora View Post
    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 \}.
    Last edited by NonCommAlg; December 6th 2008 at 05:15 PM.
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  7. #7
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    Man, you're right!
    hehe, my friend will hate me tomorow

    Thank you!!!!
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    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
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  9. #9
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    Quote Originally Posted by aurora View Post
    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.
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    I thank you so much.
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