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Math Help - Quotient Ring

  1. #1
    Senior Member vincisonfire's Avatar
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    Quotient Ring

    Just to make sure I get it right.
     \mathbb F_5[x]/(x^2+2) is a field.
    We want to express  (x^2 + x) \cdot (x^3 + 1) - (x + 1) as a polynomial of degree less than 2.
     (x^2+2-2+x)\cdot(x^3+2 -2x+1)-(x + 1)
     (x-2) \cdot (1-2x) - (x + 1)
     4x - 2x^2-3
     4x - 2x^2-4+1
     4x +1
    Thank you!
    Also, I have to find the roots of  t^2+3
     t=\pm\sqrt{-3}=\pm\sqrt{2} but 2 is not a perfect square in  \mathbb Z/5\mathbb Z
     t^2+3 has no root.
    Is that right????
    Last edited by vincisonfire; November 7th 2008 at 04:08 PM.
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  2. #2
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    Correct.

    Also note that in your field t^2+3 reduces to just 1.
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  3. #3
    Senior Member vincisonfire's Avatar
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    But  t^2 +3 reduce to 1 only if t=x. Therefore there might be roots even if x is not a root. For example, in this field,  t^2 +3 has two roots that are x and -x.
    Yes it's a cheap example but it's late and it still gives the idea. x is not a variable anymore in this field. 'Polynomials' becomes roots.
    Anyway, good night.
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    Quote Originally Posted by vincisonfire View Post
    Thank you!
    Also, I have to find the roots of  t^2+3
     t=\pm\sqrt{-3}=\pm\sqrt{2} but 2 is not a perfect square in  \mathbb Z/5\mathbb Z
     t^2+3 has no root.
    Is that right????
    The field \mathbb{F}_5 can be identified as a subfield of \mathbb{F}_5/(x^2+2).
    If \alpha = x + (x^2+2) then \alpha^2 + 2 = 0 \implies \alpha^2 = 3

    The polynomial t^2 + 3 \in \mathbb{F}_5[t] has a root in \mathbb{F}_5/(x^2+2). Consider \pm 2\alpha. Then ( \pm 2 \alpha)^2 = 4 \alpha^2 = 4\cdot 3 = 2. Therefore, (\pm 2 \alpha)^2 + 3 = 0. Which means 2\alpha,-2\alpha are its roots.
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  5. #5
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    Another (similar) way is to see that the discriminant b^2-4ac=3 is a square in the field since x^2+2=0 --> x^2=3

    Then I used the usual formula for quadratic functions and got t = (\pm\sqrt{3})/2 = \pm x * 3 (since 1/2 = 3 in Z/5Z), which is the same answer as TPH since \pm 3 = \pm 2 in this field
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