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Math Help - Prove or disprove the following set is a field.

  1. #1
    Senior Member Pinkk's Avatar
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    Prove or disprove the following set is a field.

    For a prime p, let \mathbb{Z}_{p} be the set of all equivalence classes of the relation defined by:

    aRb \Leftrightarrow a \equiv b(mod p).

    Prove or disprove: \mathbb{Z}_{p} is a field.

    So I already showed that the set has addition and multiplication well-defined. The only thing I'm having trouble with is showing that for any a \in \mathbb{Z}_{p} there exists a^{-1} \in \mathbb{Z}_{p} such that aa^{-1} = 1.

    How would I go about showing that exactly?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Hint : show that the map \pi : \mathbb{Z}_p \rightarrow \mathbb{Z}_p : x \mapsto ax is an injection, and hence a bijection. What is the inverse image of 1?
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  3. #3
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    For any a\ne0, of course...

    This is because for any two positive natural numbers m and n there exist integers a and b such that am+bn=\gcd(m,n), where \gcd(m,n) is the greatest common divisor of m and n. This fact is proved using Euclid's algorithm for finding the GCD.

    Now for any 0<n<p, \gcd(n,p)={}..., so...
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  4. #4
    Senior Member Pinkk's Avatar
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    Hmm, thanks, but someone else told me about Fermat's little theorem, which I think I'll use. But again, thanks!
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Sure, but what's the point of using another theorem if you can do without it? Can you prove Fermat's little theorem?
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  6. #6
    Senior Member Pinkk's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Sure, but what's the point of using another theorem if you can do without it? Can you prove Fermat's little theorem?
    Not really, but it makes more sense to me than anything else; I am really struggling with this material. I do not understand how to show that the suggestion you provided is a bijection, let alone how it shows that there exists an inverse.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Not really, but it makes more sense to me than anything else; I am really struggling with this material. I do not understand how to show that the suggestion you provided is a bijection, let alone how it shows that there exists an inverse.
    Use what emarakov said! Asking whether ax\equiv 1\text{ mod }n is solvable is exactly the same as asking whether the linear Diophantine equation ax+ny=1 is solvable, and as he stated this is precisely solvable when (a,n)=1. But! (a,p)=1 for any 0<a<p if p is prime
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  8. #8
    Senior Member Pinkk's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Use what emarakov said! Asking whether ax\equiv 1\text{ mod }n is solvable is exactly the same as asking whether the linear Diophantine equation ax+ny=1 is solvable, and as he stated this is precisely solvable when (a,n)=1. But! (a,p)=1 for any 0<a<p if p is prime
    Okay, this makes sense now, but isn't that another statement I'd have to prove? Okay, so using Bruno's suggestion:

    f([x]) = f([y])
    [ax] = [ay]
    [x]=[y]

    I'm not sure how to show this function is surjective and how this leads to there being an inverse element.
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  9. #9
    MHF Contributor Bruno J.'s Avatar
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    Well you should recall that if S is a finite set, and you have a function f:S\rightarrow S, then the following are equivalent :

    1. f is an injection;
    2. f is a surjection;
    3. f is a bijection.

    So certainly if \pi : \mathbb{Z}_p \rightarrow \mathbb{Z}_p is injective, then it is surjective and there exists b \in \mathbb{Z}_p such that \pi(b)=ab=1.

    So we show it's an injection.

    Suppose \pi(x_1)=\pi(x_2). Then ax_1\equiv ax_2 \mod p , so a(x_1-x_2)\equiv 0 \mod p. Now recall that this means p | a(x_1-x_2), and that since p is prime we must have either p | a or p|(x_1-x_2). By hypothesis, we have p \nmid a, therefore p |(x_1-x_2) and hence x_1 \equiv x_2 \mod p, so \pi is indeed an injection.
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  10. #10
    Senior Member Pinkk's Avatar
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    Bah, now I remember. Thanks again!
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  11. #11
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Pinkk View Post
    Bah, now I remember. Thanks again!
    You are welcome.

    In my opinion it is much better to rely on as few theorems as you can, unless you have a very good understanding of them. A friend of mine who is a PhD student once gave me a good example, after I supplied him with a proof which he found too technical (and he was right). He said : "I'll prove to you that the cube root of 2 is irrational. By Fermat's Last Theorem, a^3+a^3=b^3 has no solution in integers."
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