1. ## Degree of extension.

I have the polynomial
$
f(x) = x^5 -5x^3+5x-18
$

over $\mathbb Q$.

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

2. Originally Posted by GaloisGroup
I have the polynomial
$
f(x) = x^5 -5x^3+5x-18
$

over $\mathbb Q$.

Let $L$ be the splitting field of it.
How can I show that $[L:\mathbb Q]$ is divisible by 5.
This polyomial is irreducible. If $\alpha$ is a root of this polynomial then $\mathbb{Q}\subseteq \mathbb{Q}(\alpha) \subseteq L$. Therefore, $[L:\mathbb{Q}] = [L:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$. However, $[\mathbb{Q}(\alpha) : \mathbb{Q})] = 5$ because that polynomial is the minimimal polynomial for $\alpha$. Thus, $[L:\mathbb{Q}]$ is divisible by $5$.

In fact, $[L:\mathbb{Q}]=20$ but that takes more work to show.

3. Hi ThePerfectHacker,
how do you prove the irreducibility of that polynomial?

4. Hello,
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

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

6. Originally Posted by clic-clac
Hi moo,
looking for rational solutions would have given the answer for a polynomial in $\mathbb{Q}[X]$ of degree 2 or 3 (in such case, "no solution $\Leftrightarrow$ 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 !

7. 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 $f(x) = (x^2+ax+b)(x^3+cx^2+dx+e)$ where $a,b,c,d,e\in \mathbb{Z}$. 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 $x$ by $x+3$ and then $f(x+3) = x^5+15x^4+85x^3+225x^2+275x+105$. Now apply Einsteinstein irreducibility test with $p=5$.

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.

Stupid me >< I always miss that !

8. Hopefully there was another proof than the painful one... Nice!

9. The easiest way to show that f is irreducible is by using Eisenstein's Criterion at f(x-2) with p=5.