Field extensions and roots of polynomials

[mathgirl1]

Let F be a field extension of Q (the rationals) with [F:Q] = 24. Prove that the polynomial \(\displaystyle x^5+2x^4-16x^3+6x-10\) has no roots in F.

Proof:

Let \(\displaystyle a\) be a root of \(\displaystyle x^5+2x^4-16x^3+6x-10\). Since the polynomial has degree 5 by theorem we know that \(\displaystyle [Q(a):Q]=5\). If \(\displaystyle a \in F\) and \(\displaystyle [F:Q]=24\) then by theorem we have that \(\displaystyle [F:Q] = [F:Q(a)][Q(a):Q] \implies 24 = [F:Q(a)] 5 \) which means that \(\displaystyle [F:Q(a)] \)can not be an integer which would imply that the polynomial has not roots in F.

I think this is pretty accurate but also seems kind of too simple. Can someone please confirm whether this is correct or give advice to proceed correctly?

Thank you!

 

[Euge]

Hi mathgirl,

It's not true that since the polynomial has degree $5$, then $[\Bbb Q(a):\Bbb Q] = 5$ as a direct consequence. What you've missed in your argument is that the polynonmial is irreducible over $\Bbb Q$. Since the polynomial is irreducible of degree $5$, then $[\Bbb Q(a): \Bbb Q] = 5$. The irreducibility may be proven by applying Eisenstein's criterion for the prime $p = 2$.

 

[mathgirl1]

Ah ha! Yes! Thank you very much! I knew I was missing something. Much appreciated!