Prove for any that if is irreducible then is a square in .

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- Dec 10th 2008, 06:56 AMStigerpolynomial in finite field
Prove for any that if is irreducible then is a square in .

- Dec 12th 2008, 05:58 AMNonCommAlg
- Dec 12th 2008, 08:48 AMThePerfectHacker
Here is a more general theorem.

Let be a field of charachteristic not and be an irreducible seperable polynomial with the splitting field of over with degree . Let be roots of in (labeled in some order), define . Any can be regarded as an element of . An automorphism is even if and only if and is odd if and only if . Thus, we see from here that . Therefore, because is a Galois extension.

The quantity is known as the__discriminant__of . As**NonCommAlg**said it can be shown that the discriminant of is . But your problem asks to show that is not just in but it is in (the squares of ). To show this notice that if then for each and so . Therefore, and we see from here that .

Now if then and the Galois extension of finite fields is cyclic which means that . The only subgroup of which has order is which means that and it follows by the result above that the discriminant is a square.