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

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

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

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

    attempt:

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

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

    (c) $\displaystyle remainder\;of\;\frac{3\cdot(x)}{7}=1$
    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 $\displaystyle \mathbb{Z}_m$ is a field for $\displaystyle m$ prime, you just need to show that it is an integral domain.

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

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

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

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

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

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

    (c) $\displaystyle remainder\;of\;\frac{3\cdot(x)}{7}=1$
    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 $\displaystyle \mathbb{Z}_m$ is a field for $\displaystyle m$ prime, you just need to show that it is an integral domain.

    For the other direction, assume $\displaystyle m$ is not prime. Then we have $\displaystyle m=nr$, where $\displaystyle 1<n,r<m$. But $\displaystyle nr=0$ in $\displaystyle \mathbb{Z}_m$, which contradicts the fact that $\displaystyle \mathbb{Z}_m$ 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 $\displaystyle Z_m$ 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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