# Degree of extension.

• Apr 1st 2009, 02:21 AM
GaloisGroup
Degree of extension.
I have the polynomial
$\displaystyle f(x) = x^5 -5x^3+5x-18$
over $\displaystyle \mathbb Q$.

Let $\displaystyle L$ be the splitting field of it.
How can I show that $\displaystyle [L:\mathbb Q]$ is divisible by 5.
• Apr 2nd 2009, 06:54 PM
ThePerfectHacker
Quote:

Originally Posted by GaloisGroup
I have the polynomial
$\displaystyle f(x) = x^5 -5x^3+5x-18$
over $\displaystyle \mathbb Q$.

Let $\displaystyle L$ be the splitting field of it.
How can I show that $\displaystyle [L:\mathbb Q]$ is divisible by 5.

This polyomial is irreducible. If $\displaystyle \alpha$ is a root of this polynomial then $\displaystyle \mathbb{Q}\subseteq \mathbb{Q}(\alpha) \subseteq L$. Therefore, $\displaystyle [L:\mathbb{Q}] = [L:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$. However, $\displaystyle [\mathbb{Q}(\alpha) : \mathbb{Q})] = 5$ because that polynomial is the minimimal polynomial for $\displaystyle \alpha$. Thus, $\displaystyle [L:\mathbb{Q}]$ is divisible by $\displaystyle 5$.

In fact, $\displaystyle [L:\mathbb{Q}]=20$ but that takes more work to show.
• Apr 3rd 2009, 07:35 AM
clic-clac
Hi ThePerfectHacker,
how do you prove the irreducibility of that polynomial?
• Apr 3rd 2009, 09:43 AM
Moo
Hello,
Quote:

Originally Posted by clic-clac
Hi ThePerfectHacker,
how do you prove the irreducibility of that polynomial?

I know nothing of Galois theory, so I'm not too sure you're looking for rational solutions, real solutions, or complex solutions... (my guess goes for rational zeroes ^^)

Anyway, this may help : The Rational Roots Test
• Apr 3rd 2009, 10:31 AM
clic-clac
Hi moo,
looking for rational solutions would have given the answer for a polynomial in $\displaystyle \mathbb{Q}[X]$ of degree 2 or 3 (in such case, "no solution $\displaystyle \Leftrightarrow$ irreducibility"). With higher degrees, the polynomial can be a product of irreducible polynomials and be reducible without having ant root. But thanks :)
• Apr 3rd 2009, 10:34 AM
Moo
Quote:

Originally Posted by clic-clac
Hi moo,
looking for rational solutions would have given the answer for a polynomial in $\displaystyle \mathbb{Q}[X]$ of degree 2 or 3 (in such case, "no solution $\displaystyle \Leftrightarrow$ irreducibility"). With higher degrees, the polynomial can be a product of irreducible polynomials and be reducible without having ant root. But thanks :)

Stupid me >< I always miss that !
• Apr 3rd 2009, 10:37 AM
ThePerfectHacker
Quote:

Originally Posted by clic-clac
Hi ThePerfectHacker,
how do you prove the irreducibility of that polynomial?

If you use the rational zeros theorem you will realize it has no zeros. Therefore, if this polynomial is reducible then it means $\displaystyle f(x) = (x^2+ax+b)(x^3+cx^2+dx+e)$ where $\displaystyle a,b,c,d,e\in \mathbb{Z}$. Expand RHS and compare coefficients and show that is impossible. This is a really painful way of proving irreducibility! Here is an easier way. Replace $\displaystyle x$ by $\displaystyle x+3$ and then $\displaystyle f(x+3) = x^5+15x^4+85x^3+225x^2+275x+105$. Now apply Einsteinstein irreducibility test with $\displaystyle p=5$. (Nod)

Quote:

Originally Posted by Moo Princess
Hello,

I know nothing of Galois theory, so I'm not too sure you're looking for rational solutions, real solutions, or complex solutions... (my guess goes for rational zeroes ^^)

Anyway, this may help : The Rational Roots Test

I do not want to be insulting, but how many times do I have to tell you that not having zeros does not imply irreducibility! I think that is my third time telling you this already. (Smile)

Quote:

Stupid me >< I always miss that !
(Rofl)
• Apr 3rd 2009, 10:50 AM
clic-clac
Hopefully there was another proof than the painful one...:) Nice!
• Apr 3rd 2009, 11:28 AM
GaloisGroup
The easiest way to show that f is irreducible is by using Eisenstein's Criterion at f(x-2) with p=5.