It is currently Wed May 23, 2018 7:15 pm


All times are UTC [ DST ]




Post new topic Reply to topic  [ 2 posts ] 
Author Message
 Post subject: Roots of a polynomial
PostPosted: Thu Jun 09, 2016 8:29 am 
Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
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.


Top
Offline Profile  
Reply with quote  

PostPosted: Thu Jun 09, 2016 8:30 am 

Joined: Mon Nov 09, 2015 11:52 am
Posts: 76
Location: Limassol/Pyla Cyprus
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.


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 2 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net