I have the polynomial over . It can be easily seen that Klein's Four group is the Galois group of this polynomial.

How can I show that the corresponding polynomial

is irreducible in for every prime .

Printable View

- Mar 24th 2009, 12:51 PMZetaXPolynomial
I have the polynomial over . It can be easily seen that Klein's Four group is the Galois group of this polynomial.

How can I show that the corresponding polynomial

is irreducible in for every prime . - Mar 24th 2009, 01:06 PMMoo
Oh man, please, it's polynom

**ia**l ! - Mar 27th 2009, 12:26 PMThePerfectHacker
This is not true. If you take , . I think you want to prove that this polynomial is

**always reducible**as opposed to irreducible. One way to proved is by Dedekind's theorem (again) and by an understanding of transitive subgroups of . The transitive subgroup (Klein four group) consists of an identity element and -cycle products ( ). Therefore, mod any the polynomial either splits or factors into quadradic factors. Therefore, it is always reducible. - Mar 28th 2009, 01:17 AMZetaX
Sorry what I meant was reducible not irreducible, it was a mistake.

- Mar 28th 2009, 03:53 PMThePerfectHacker
Here is another way of proving this which is elementary. Let be prime with .

Notice the following factorizations:

[1]

[2]

[3]

If is a square root in then we can use the factorization in [1] where we replace by a square root of modulo . If is a square root in then we can use the factorization in [2] where we replace by a square root of modulo . If neither are squares mod then it means (Legendre symbol). However, this means . Therefore, would be a square in and so we can use the factorization in [3].

Therefore, is always reducible for any . - Mar 29th 2009, 11:04 AMZetaX
why should p be greater than or equal to 7?

- Mar 29th 2009, 11:10 AMThePerfectHacker