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Roots of a polynomial

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

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

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.