Roots of a polynomial

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
Post Reply
Tsakanikas Nickos
Community Team
Posts: 314
Joined: Tue Nov 10, 2015 8:25 pm

Roots of a polynomial

#1

Post by Tsakanikas Nickos »

Let \( \displaystyle f(x) \in \mathbb{Q}[x] \) be an irreducible polynomial over \( \displaystyle \mathbb{Q} \) and let \( \displaystyle \mathbb{K} \) be its splitting field over \( \displaystyle \mathbb{Q}. \) Suppose that the degree \( \displaystyle \left[ \mathbb{K} : \mathbb{Q} \right] \) of the extension \( \displaystyle \mathbb{K} / \mathbb{Q} \) is odd. Show that all roots of \( \displaystyle f(x) \) are real.
Demetres
Former Team Member
Former Team Member
Posts: 77
Joined: Mon Nov 09, 2015 11:52 am
Location: Limassol/Pyla Cyprus
Contact:

Re: Roots of a polynomial

#2

Post by Demetres »

We observe that complex conjugation is a field automorphism of \(\mathbb{K}\). If \(f(x)\) has a non-real root, then \(\mathbb{K}\) has a non-real element and therefore this automorphism has order \(2\). Let \(\mathbb{L}\) be the fixed field of the automorphism. (I.e. \(\mathbb{L}=\mathbb{K} \cap \mathbb{R}\).) By the fundamental theorem of Galois theory we have that \([\mathbb{K}:\mathbb{L}]=2\) and so \([\mathbb{K}:\mathbb{Q}]=[\mathbb{K}:\mathbb{L}][\mathbb{L}:\mathbb{Q}]\) is even, a contradiction.
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 8 guests