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Roots of a polynomial

Linear Algebra, Algebraic structures (Groups, Rings, Modules, etc), Galois theory, Homological Algebra
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Tsakanikas Nickos
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Roots of a polynomial

#1

Post by Tsakanikas Nickos » Thu Jun 09, 2016 8:29 am

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
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Re: Roots of a polynomial

#2

Post by Demetres » Thu Jun 09, 2016 8:30 am

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.
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