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Math Help - Prove that for every prime p, there is an irreducible quadratic in Z_p[x]

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    Junior Member Dark Sun's Avatar
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    Prove that for every prime p, there is an irreducible quadratic in Z_p[x]

    Prove that for every prime p, there is an irreducible quadratic in Z_p[x] (the polynomial ring of the integers mod p).

    I have not been able to wrap my brain around this one. Does anyone have any ideas?

    I have been looking at x^2+x+1 and x^2+1, but haven't been able to derive a contradiction.

    This is what I have so far: Assume x^2+1 is reducible in Z_p[x]. Then, \exists a,b\in Z_p[x] s.t. x^2+1=(x-a)(x-b)=x^2-(a+b)x+ab\Rightarrow a+b=0, ab=1

    Thanks!
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    Quote Originally Posted by Dark Sun View Post
    Prove that for every prime p, there is an irreducible quadratic in Z_p[x] (the polynomial ring of the integers mod p).

    I have not been able to wrap my brain around this one. Does anyone have any ideas?

    I have been looking at x^2+x+1 and x^2+1, but haven't been able to derive a contradiction.

    This is what I have so far: Assume x^2+1 is reducible in Z_p[x]. Then, \exists a,b\in Z_p[x] s.t. x^2+1=(x-a)(x-b)=x^2-(a+b)x+ab\Rightarrow a+b=0, ab=1

    Thanks!
    if p = 2, consider x^2 + x + 1. for p > 2, choose a \in \{1, \cdots, p-1 \} such that a is not a quadratic residue modulo p. then x^2 - a would be irreducible in \mathbb{Z}_p[x].
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    Junior Member Dark Sun's Avatar
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    Thanks Dark Knight, this is just the answer I was looking for.

    You have helped me a great deal.

    I wish you well in your studies!
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    Quote Originally Posted by Dark Sun View Post
    Prove that for every prime p, there is an irreducible quadratic in Z_p[x] (the polynomial ring of the integers mod p).
    Here is another approach but this is not a good approach at all.

    Let f(x) = x^{p^2} - x and let K be its splitting field.
    Then we have that [K:F]=2, pick any a\in K - F then the minimal polynomial of a must have degree 2.
    Thus, you have found an irreducible polynomial of degree 2.
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