# polynomial in finite field

• Dec 10th 2008, 06:56 AM
Stiger
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 $\mathbb{F}_{p^n}$.
• Dec 12th 2008, 05:58 AM
NonCommAlg
Quote:

Originally Posted by Stiger
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 $\delta \in \mathbb{F}_{p^n}. \ \Box$
• Dec 12th 2008, 08:48 AM
ThePerfectHacker
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. 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.