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Thread: Quotient Ring

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

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

    Also note that in your field $\displaystyle t^2+3$ reduces to just $\displaystyle 1$.
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  3. #3
    Senior Member vincisonfire's Avatar
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    But $\displaystyle 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, $\displaystyle 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 $\displaystyle t^2+3$
    $\displaystyle t=\pm\sqrt{-3}=\pm\sqrt{2}$ but 2 is not a perfect square in $\displaystyle \mathbb Z/5\mathbb Z$
    $\displaystyle t^2+3$ has no root.
    Is that right????
    The field $\displaystyle \mathbb{F}_5$ can be identified as a subfield of $\displaystyle \mathbb{F}_5/(x^2+2)$.
    If $\displaystyle \alpha = x + (x^2+2)$ then $\displaystyle \alpha^2 + 2 = 0 \implies \alpha^2 = 3$

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

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