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Math Help - Irreducible Polynomial

  1. #1
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    Irreducible Polynomial

    Is the given polynomial irreducible:

    (x^2)+x-2 in Z3, Z7?

    Thanks. MK.
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  2. #2
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    Quote Originally Posted by MKLyon View Post
    Is the given polynomial irreducible:

    (x^2)+x-2 in Z3, Z7?

    Thanks. MK.
    Whenever you a given a polynomial over a field of degree three or two simply look if it has zero's to determine if it is reducible.

    Check 0,1,2\in \mathbb{Z}_3[x]
    We have,
    0^2+0-2=1
    1^2+1-1=1
    2^2+2-1=1+2-1=2
    Thus, it has no zeros.
    Thus it is irreducible over \mathbb{Z}_3[x].

    Do the same for \mathbb{Z}_7[x]
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    Why does it turn into -1 instead of -2?

    Thanks for the earlier response.
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    Quote Originally Posted by MKLyon View Post
    Why does it turn into -1 instead of -2?
    You mean why does it turn into 1 instead of -2.
    Because -2 is defined as the additive inverse in the field \mathbb{Z}_3[x] the additive inverse of 2 is 1 because 1+2=0 (we are in the set {0,1,2} definied addition and multiplication modulo 3).
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    By this, I concluded that the equation was reducible in Z7[x] because 3 and 6 produced zeros. Can anyone confirm this?
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  6. #6
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    Quote Originally Posted by ThePerfectHacker View Post
    Whenever you a given a polynomial over a field of degree three or two simply look if it has zero's to determine if it is reducible.

    Check 0,1,2\in \mathbb{Z}_3[x]
    We have,
    0^2+0-2=1
    -1=1" alt="1^2+1-1=1" />
    -1=1+2-1=2" alt="2^2+2-1=1+2-1=2" />
    Thus, it has no zeros.
    Thus it is irreducible over \mathbb{Z}_3[x].

    Do the same for \mathbb{Z}_7[x]
    Quote Originally Posted by MKLyon View Post
    Why does it turn into -1 instead of -2?

    Thanks for the earlier response.
    I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:
    1^2+1-2=0
    2^2+2-2=1+2-2=1

    which shows that the polynomial IS reducible in Z3[x].

    (Okay, so the red coloring didn't work in the quote! You still get the idea.)

    -Dan

    PS: That x = 1 produces x^2+x-2 = 0 implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that x^2 + x - 2 = (x - 1)(x + 2). Where is the x + 2 factor in the above list? Well, this would imply a root of x = -2 = 1 (mod 3), so really x = 1 again. This says that:
    x^2 + x - 2 = (x - 1)(x + 2) = (x - 1)^2 in Z3[x].
    Last edited by topsquark; October 22nd 2006 at 08:16 AM. Reason: Coding failure?
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by MKLyon View Post
    By this, I concluded that the equation was reducible in Z7[x] because 3 and 6 produced zeros. Can anyone confirm this?
    Using TPH's method I got that
    1^2 + 1 - 2 = 0
    5^2 + 5 - 2 = 0

    So yes, it should be reducible.

    -Dan
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  8. #8
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    Quote Originally Posted by topsquark View Post
    I believe MKLyon was talking about these -1s. He is correct that there is a typo. The lines should read:
    1^2+1-2=0
    2^2+2-2=1+2-2=1

    which shows that the polynomial IS reducible in Z3[x].

    (Okay, so the red coloring didn't work in the quote! You still get the idea.)

    -Dan

    PS: That x = 1 produces x^2+x-2 = 0 implies that x - 1 is a factor of the polynomial in Z3[x]. By doing the long division you can show that x^2 + x - 2 = (x - 1)(x + 2). Where is the x + 2 factor in the above list? Well, this would imply a root of x = -2 = 1 (mod 3), so really x = 1 again. This says that:
    x^2 + x - 2 = (x - 1)(x + 2) = (x - 1)^2 in Z3[x].
    I see what I did wrong now,, thank you.
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