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Math Help - Prove that Z_m is a field iff m is a prime.

  1. #1
    Member Mollier's Avatar
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    Prove that Z_m is a field iff m is a prime.

    Hi,

    problem:
    Let m be an integer, m\geq2 and let \mathbb{Z}_m be the set of all positive integers less than m, \mathbb{Z}_m=\{0,1,\cdots,m-1\}.
    If \alpha and \beta are in \mathbb{Z}_m, let \alpha+\beta be the least positive remainder obtained by dividing the (ordinary) sum of \alpha and \beta by m,
    and similarly, let \alpha\beta be the least positive remainder obtained by dividing the (ordinary) product of \alpha and \beta by m.

    (a) Prove that \mathbb{Z}_m is a field if and only if m is a prime.

    (b) What is -1 in \mathbb{Z}_5?

    (c) What is \frac{1}{3} in \mathbb{Z}_7?

    attempt:

    (a) I could really use a hint or two on this one..

    (b) remainder\;of\;\frac{1+(x)}{5}=0
    Answer: 4.

    (c) remainder\;of\;\frac{3\cdot(x)}{7}=1
    Answer:5.

    Btw, is there a nice math symbol for "the remainder of"?

    Thanks.
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  2. #2
    Member Black's Avatar
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    Finite integral domains are fields, so to show that \mathbb{Z}_m is a field for m prime, you just need to show that it is an integral domain.

    For the other direction, assume m is not prime. Then we have m=nr, where 1<n,r<m. But nr=0 in \mathbb{Z}_m, which contradicts the fact that \mathbb{Z}_m is an integral domain.
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  3. #3
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    Quote Originally Posted by Mollier View Post
    Hi,

    problem:
    Let m be an integer, m\geq2 and let \mathbb{Z}_m be the set of all positive integers less than m, \mathbb{Z}_m=\{0,1,\cdots,m-1\}.
    If \alpha and \beta are in \mathbb{Z}_m, let \alpha+\beta be the least positive remainder obtained by dividing the (ordinary) sum of \alpha and \beta by m,
    and similarly, let \alpha\beta be the least positive remainder obtained by dividing the (ordinary) product of \alpha and \beta by m.

    (a) Prove that \mathbb{Z}_m is a field if and only if m is a prime.

    (b) What is -1 in \mathbb{Z}_5?

    (c) What is \frac{1}{3} in \mathbb{Z}_7?

    attempt:

    (a) I could really use a hint or two on this one..
    Use the fact that a field has no 0 divisors. That is, there are no members, x and y such that xy= 0. Now prove the contrapositive of each direction.

    To prove "If Z_m is a field the m is prime" start "if m is NOT prime, there exist a, b< m such that ab= m" and show that there are 0 divisors.

    To prove "if m is prime, then Z_m is a field" start "if Z_m is not a field, then there must exist zero divisors- a, b< m such that ab= 0 (mod m)" and show that m= ab.

    (b) remainder\;of\;\frac{1+(x)}{5}=0
    Answer: 4.

    (c) remainder\;of\;\frac{3\cdot(x)}{7}=1
    Answer:5.

    Btw, is there a nice math symbol for "the remainder of"?

    Thanks.
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  4. #4
    Member Mollier's Avatar
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    Quote Originally Posted by Black View Post
    Finite integral domains are fields, so to show that \mathbb{Z}_m is a field for m prime, you just need to show that it is an integral domain.

    For the other direction, assume m is not prime. Then we have m=nr, where 1<n,r<m. But nr=0 in \mathbb{Z}_m, which contradicts the fact that \mathbb{Z}_m is an integral domain.
    I understand your explanation now that HallsOfIvy dumbed it down for me.
    Thanks Black!

    Quote Originally Posted by HallsofIvy View Post
    Use the fact that a field has no 0 divisors. That is, there are no members, x and y such that xy= 0. Now prove the contrapositive of each direction.

    To prove "If Z_m is a field the m is prime" start "if m is NOT prime, there exist a, b< m such that ab= m" and show that there are 0 divisors.

    To prove "if m is prime, then Z_m is a field" start "if Z_m is not a field, then there must exist zero divisors- a, b< m such that ab= 0 (mod m)" and show that m= ab.
    Very nice, thank you very much!
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  5. #5
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    Quote Originally Posted by Mollier View Post
    I understand your explanation now that HallsOfIvy dumbed it down for me.
    There is nothing I am better at that "dumbing down"!
    ("dumb" just comes naturally to me.)

    Thanks Black!



    Very nice, thank you very much!
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  6. #6
    Member Mollier's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    There is nothing I am better at that "dumbing down"!
    ("dumb" just comes naturally to me.)
    Dumb seems to be a common talent.
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