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Math Help - polynomial in finite field

  1. #1
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    polynomial in finite field

    Prove for any a, b \in \mathbb{F}_{p^n} that if x^3+ax+b is irreducible then -4a^3-27b^2 is a square in \mathbb{F}_{p^n}.
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  2. #2
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    Quote Originally Posted by Stiger View Post
    Prove for any a, b \in \mathbb{F}_{p^n} that if x^3+ax+b is irreducible then -4a^3-27b^2 is a square in \mathbb{F}_{p^n}.
    the splitting field of p(x)=x^3 + ax +b is \mathbb{F}_q, where q=p^{3n}. let \alpha, \beta, \gamma be the roots of p(x). a simple calculation shows that -4a^3-27b^2=[(\alpha-\beta)(\beta - \gamma)(\gamma - \alpha)]^2.

    now \delta=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) is invariant under any \mathbb{F}_{p^n}-automorphism of \mathbb{F}_q. (why? be careful here!) thus \delta \in \mathbb{F}_{p^n}. \ \Box
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  3. #3
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    Here is a more general theorem.

    Let F be a field of charachteristic not 2 and f(x)\in F[x] be an irreducible seperable polynomial with K the splitting field of f(x) over F with degree n. Let \alpha_1,\alpha_2,...,\alpha_n be roots of f(x) in K (labeled in some order), define \Delta = \prod_{i<j}(\alpha_i - \alpha_j). Any \sigma \in \text{Gal}(K/F) can be regarded as an element of S_n. An automorphism \sigma is even if and only if \sigma (\Delta) =  \Delta and is odd if and only if \sigma (\Delta) = - \Delta. Thus, we see from here that \sigma (\Delta^2) = \Delta^2. Therefore, \Delta^2 \in K^{\text{Gal}(K/F)} = F because K/F is a Galois extension.

    The quantity \Delta^2 is known as the discriminant of f(x). As NonCommAlg said it can be shown that the discriminant of x^3 + ax + b is \Delta^2 = - 4a^3 - 27b^2. But your problem asks to show that \Delta^2 is not just in F but it is in F^2 (the squares of F). To show this notice that if \text{Gal}(K/F)\subseteq A_n then \sigma (\Delta) = \Delta for each \sigma \in \text{Gal}(K/F) and so \Delta \in K^{\text{Gal}(K/F)} = F. Therefore, \Delta \in F \implies \Delta^2 \in F^2 and we see from here that -4a^3 - 27b^2 = \Delta^2 \in F^2.

    Now if F = \mathbb{F}_q then K = \mathbb{F}_{q^3} and the Galois extension of finite fields is cyclic which means that G=\text{Gal}(K/F) \simeq \mathbb{Z}_3. The only subgroup of S_3 which has order 3 is A_3 which means that G \subseteq A_3 and it follows by the result above that the discriminant is a square.
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