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.

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- Apr 1st 2009, 02:21 AMGaloisGroupDegree 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 PMThePerfectHacker
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 AMclic-clac
Hi ThePerfectHacker,

how do you prove the irreducibility of that polynomial? - Apr 3rd 2009, 09:43 AMMoo
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 - Apr 3rd 2009, 10:31 AMclic-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 AMMoo
- Apr 3rd 2009, 10:37 AMThePerfectHacker
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)

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 !

- Apr 3rd 2009, 10:50 AMclic-clac
Hopefully there was another proof than the painful one...:) Nice!

- Apr 3rd 2009, 11:28 AMGaloisGroup
The easiest way to show that f is irreducible is by using Eisenstein's Criterion at f(x-2) with p=5.