# irreducibility

• Apr 21st 2009, 06:04 PM
dori1123
irreducibility
Find a primitive generator for $\displaystyle \mathbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ over $\displaystyle \mathbb{Q}$.
I found the polynomial for the root $\displaystyle \alpha=\sqrt2+\sqrt3 +\sqrt5$, which is $\displaystyle x^8-40x^6+352x^4-960x^2+576$. But I'm having trouble showing the polynomial is irreducible. Some help please.
• Apr 21st 2009, 08:42 PM
Gamma
Here is an idea
I am not sure how to show immediately that it is irreducible; however here is another idea.

We know that:
$\displaystyle [\mathbb{Q}(\sqrt2 , \sqrt3, \sqrt5):\mathbb{Q}]=[\mathbb{Q}(\sqrt2 , \sqrt3, \sqrt5):\mathbb{Q}(\sqrt2 , \sqrt3)][\mathbb{Q}(\sqrt2 , \sqrt3):\mathbb{Q}(\sqrt2)][\mathbb{Q}(\sqrt2):\mathbb{Q}]$

But it is easy to show each of the intermediary steps is a degree 2 extension with irreducible minimal polynomials $\displaystyle x^2-5, x^2-3, x^2 - 2$ all irreducible by Eisenstein with p = 5, 3, 2 respectively. So that means the total extension is:
$\displaystyle [\mathbb{Q}(\sqrt2 , \sqrt3, \sqrt5):\mathbb{Q}]=2*2*2=8$
Since you have found a monic polynomial of degree 8 that has this as a root it must be irreducible. Kind of roundabout, but hope it helps.