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Thread: Roots

  1. #1
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    Roots

    Find the roots of x^3 - 1 in F (sub 7)

    Find the roots of x^3 - 1 in F (sub 5)

    (There is a bar over the 1's in the above problems)
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by jzellt View Post
    Find the roots of x^3 - 1 in F (sub 7)

    Find the roots of x^3 - 1 in F (sub 5)

    (There is a bar over the 1's in the above problems)
    Hi jzellt.

    5 and 7 are pretty small numbers. You can simply take the cubes of all the small numbers and note which ones are congruent to 1 modulo 5 / 7.
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  3. #3
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    Can you be a little more specific?

    Maybe work out the first one...I'm really lost with this topic.
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  4. #4
    Super Member Gamma's Avatar
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    A polynomial of degree n has at most n roots.

    There are only n elements in $\displaystyle \mathbb{Z}_n$. Try them is what he is saying.

    $\displaystyle 0^3-1=0-1=-1 \equiv 4 $ (mod 5)
    $\displaystyle 1^3-1=1-1=0 \equiv 0 $ (mod 5)
    $\displaystyle 2^3-1=8-1=7 \equiv 2 $ (mod 5)
    $\displaystyle 3^3-1=27-1=26 \equiv 1 $ (mod 5)
    $\displaystyle 4^3-1=64-1=63 \equiv 3 $ (mod 5)

    Thus the only root is 1.

    I think you can do it for the other one.
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  5. #5
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    So, from what you did...How do I know that the only root is 1?
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  6. #6
    Super Member Gamma's Avatar
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    Do you know what a root of a polynomial is?

    It is an element of the field for which the polynomial evaluated at that element yields 0.

    I just checked every element of the field (all 5 of them) to see if the polynomial was 0 in the field or not. Only 1 was a root. That is how I knew.
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  7. #7
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    Quote Originally Posted by jzellt View Post
    So, from what you did...How do I know that the only root is 1?
    What else could it be?

    $\displaystyle 0^3-1=0-1=-1 \equiv 4 (mod 5)$ so it's not 0.
    $\displaystyle 1^3-1=1-1= 0 (mod 5)$ so 1 is a root.
    $\displaystyle 2^3-1=8-1= 7 \equiv 2 (mod 5)$ so it's not 2.
    $\displaystyle 3^3-1=27-1= 26 \equiv 1 (mod 5)$ so it's not 3.
    $\displaystyle 4^3-1=64-1= 63 \equiv 3 (mod 5)$ so it's not 4.

    There are no other possibilities.
    Last edited by HallsofIvy; May 9th 2009 at 10:36 AM.
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  8. #8
    Senior Member TheAbstractionist's Avatar
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    Okay, here is some group-theory analysis (although this particular problem is so simple that it hardly needs this kind of thing).

    Now, if $\displaystyle p$ is a prime, $\displaystyle \mathbb Z_p^\times=\{1,\ldots,p-1\}$ is a group under multiplication modulo $\displaystyle p.$ Now, $\displaystyle x^3-1$ is a root in $\displaystyle \mathbb Z_p^\times=\{1,\ldots,p-1\}$ if and only if the subgroup generated by $\displaystyle x,$ namely $\displaystyle \left<x\right>$ has order dividing 3. Since 3 is a prime, this means that either (i) $\displaystyle \left<x\right>$ is trivial or (ii) a cyclic group of order 3. If $\displaystyle p=5$ then $\displaystyle p-1=4$ and so by Lagrange there is no subgroup of order 3 and so the only solution to the equation is $\displaystyle x=1.$ But for $\displaystyle p=7$ it is possible to have to have a subgroup of order 3 and so it turns out that there are two solutions modul0 7.

    The conclusion is: $\displaystyle x^3-1$ has one root in $\displaystyle \mathbb F_5$ (namely 1) and three roots in $\displaystyle \mathbb F_7 $ (nanely 1, 2 and $\displaystyle 2^2=4.)$

    If there is anything not clear, please ask.
    Last edited by TheAbstractionist; May 10th 2009 at 12:47 AM.
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  9. #9
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    So after working both problems out...

    The only root is 1 regardless if im working in F7 or F5

    Is this correct?
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  10. #10
    Super Member Gamma's Avatar
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    Quote Originally Posted by TheAbstractionist View Post
    Okay, here is some group-theory analysis (although this particular problem is so simple that it hardly needs this kind of thing).

    Now, if $\displaystyle p$ is a prime, $\displaystyle \mathbb Z_p^\times=\{1,\ldots,p-1\}$ is a group under multiplication modulo $\displaystyle p.$ Now, $\displaystyle x^3-1$ is a root in $\displaystyle \mathbb Z_p^\times=\{1,\ldots,p-1\}$ if and only if the subgroup generated by $\displaystyle x,$ namely $\displaystyle \left<x\right>$ has order dividing 3. Since 3 is a prime, this means that either (i) $\displaystyle \left<x\right>$ is trivial or (ii) a cyclic group of order 3. If $\displaystyle p=5$ then $\displaystyle p-1=4$ and so by Lagrange there is no subgroup of order 3 and so the only solution to the equation is $\displaystyle x=1.$ But for $\displaystyle p=7$ it is possible to have to have a subgroup of order 3 and so it turns out that there are two solutions modul0 7.

    The conclusion is: $\displaystyle x^3-1$ has one root in $\displaystyle \mathbb F_5$ (namely 1) and three roots in $\displaystyle \mathbb F_7 $ (nanely 1, 2 and $\displaystyle 2^2=4.)$

    If there is anything not clear, please ask.
    For $\displaystyle \mathbb{Z}_5$ as the abstractionist and I have both shown you already 1 is indeed the only root to that polynomial.

    However, this is not the case for $\displaystyle \mathbb{Z}_7$ as theabstractionist was so kind to point out, both 2 and 4 are roots as well.

    Please take the time to read people's responses carefully, they are not answering your questions for their own health.

    $\displaystyle 1^3-1=0 \equiv 0 $ (mod 7) so 1 is a root

    $\displaystyle 2^3-1=7 \equiv 0 $ (mod 7) so 2 is a root

    $\displaystyle 4^3-1=64-1=63=7*9 \equiv 0 $ (mod 7) so 4 is a root.

    A polynomial of degree n can have at most n roots, so this is all of them as $\displaystyle x^3-1$ has degree 3 and we have listed 3 roots.

    Clear?
    Last edited by Gamma; May 12th 2009 at 11:49 PM. Reason: typo
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