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Math Help - Reducible problem. Help!

  1. #1
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    Reducible problem. Help!

    Suppose p>2 is a prime and x^2+1 is reducible in Fp[x]. Prove that the multiplicative group (Fp\{0}, \cdot) contains an element of order 4. Deduce that p \equiv1mod4.

    I'm finding algebra quite hard atm, and this question has got me stuck. I don't know where to start. Please help!
    Last edited by gizmo; April 13th 2010 at 08:48 AM.
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  2. #2
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    Quote Originally Posted by gizmo View Post
    Suppose p>2 is a prime and x^2+1 is reducible in Fp[x]. Prove that the multiplicative group (Fp\{0}, \cdot) contains an element of order 4. Deduce that p \equiv1mod4.

    I'm finding algebra quite hard atm, and this question has got me stuck. I don't know where to start. Please help!

    x^2+1 reducible mod p \Longrightarrow x^2=-1\!\!\!\pmod p has a solution in \left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}.

    Now, x^2=-1\!\!\!\pmod p\Longrightarrow x^4=1\!\!\!\pmod p\Longrightarrow x\in\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*} has order 4 \Longrightarrow 4\mid ord\left(\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}\right)|=p-1\Longrightarrow ...end the argument now.

    Tonio
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  3. #3
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    Thank you so much, that makes sense to me now! Hopefully I'l be able to make some progress through the rest of my algebra
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  4. #4
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    Quote Originally Posted by tonio View Post
    \Longrightarrow 4\mid ord\left(\left(\mathbb{Z}\slash p\mathbb{Z}\right)^{*}\right)|=p-1\Longrightarrow ...end the argument now.

    Tonio
    Sorry, I followed up to this point, but I'm a bit confused as to what this first term means?
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  5. #5
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    Quote Originally Posted by gizmo View Post
    Sorry, I followed up to this point, but I'm a bit confused as to what this first term means?


    "Four divides the order of the multiplicative group of non-zero residues modulo p, whose order is p-1"

    Tonio
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  6. #6
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    thanks! i think i get this now, so to finish it off...
    4|p-1 \Rightarrow p-1 \equiv 0mod4 \Rightarrow p \equiv 1mod4
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