Let $x$ be $\cos \displaystyle \frac \pi n$ for some natural number $n$. Then is it true that $\mathbb{Q}(x^2+x)=\mathbb{Q}(x)$?
the number field $\mathbb{Q}(\cos \frac \pi n)$
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number-theory
field-theory
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0I don't have any good idea. We can consider the field $\mathbb{Q}(cos \frac \pi n)$ as the maximal real subfield of the cyclotomic field $\mathbb{Q}(\zeta_{2n})$. – 2012-09-17
1 Answers
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Let $u=x^2+x$ then $x$ satisfies the quadratic equation $X^2+X-u$ over $Q(u)$, so $Q(x)$ is at most a quadratic extension of $Q(u)$. It is known that $\deg(x)=\phi(n)/2$; in case $n=p$ a prime then $\deg(x)=(p-1)/2$.
Thus if $p=3 \mod 4$ then $\deg(x)$ is odd so $Q(u)=Q(x)$.
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0I have to show $1+4u$ is a square in $Q(u)$. Haven't done it yet. Can you? – 2012-09-17