The subextension of the 7th cyclotomic field of degree $2$

Question

The Galois group of the 7th cyclotomic field is $Z_6$, so it has a unique subgroup of degree $3$. Let $x = e^{\frac{2\pi i }{7}}$, then it is easy to say that $a = x^4+x^2+x$ is fixed by this subgroup. My question is how do I show that $x^4+x^2+x$ is not rational. That seems like an easy question, but I do not even know where to start since I cannot envision the concrete form of $x^4+x^2+x$.

Answer 1

There is a neat trick here which I learned by accident.

Consider also the element $b=x^{-1}+x^{-2}+x^{-4}=x^6+x^5+x^3$, then $a$ and $b$ are roots of a quadratic polynomial with integer coefficients and can be determined explicitly.

In an extension of degree $2$ over the rationals any element $\alpha$ which is not rational will generate the whole extension. Since the extension is of degree $2$ the elements $1, \alpha, \alpha^2$ cannot be linearly independent over $\mathbb Q$ so $\alpha$ satisfies a quadratic polynomial. Clearing denominators gives a quadratic with integer coefficients. If $\alpha$ is an algebraic integer (and $a$ here is constructed to be such) then there will be a quadratic with leading coefficient $1$.

Note that $a+b=-1$ so that if $a$ is fixed under the various automorphisms you are considering, $b$ will be fixed too. And $b=-a-1$

Note also that the mundane procedure indicated above works here as well:

$$a^2=(x+x^2+x^4)^2=x^2+x^4+x^8+2(x^3+x^5+x^6)=a+2(-a-1)$$ so that $a^2+a+2=0$

Answer 2

More generally, if $p$ is an odd prime, then the $p$th cyclotomic field has, as its quadratic subextension, ${\bf Q}(\sqrt p)$ if $p\equiv1\bmod4$, and ${\bf Q}(\sqrt{-p})$ if $p\equiv-1\bmod4$. Even more generally, there are similar formulas for the unique quadratic subextension of the $n$th cyclotomic field. These can be worked out from the Wikipedia pages on quadratic Gauss sums, https://en.wikipedia.org/wiki/Quadratic_Gauss_sum

Answer 3

This kind of question has been asked many times and admits a general answer. To keep it at hand, let me give a detailed explanation. For a given odd prime $p$, the cyclotomic field $F=\mathbf Q(\zeta_p)$ is a cyclic extension of $\mathbf Q$ of degree $p-1$. By the Galois correspondance, it contains a unique quadratic subfield $K$ which we can make explicit. To this end, let us compute the discriminant $D(F)$ using the formula $D(F)=N(f'(\zeta_p))$ , where $f$ is the minimal plynomial of $\zeta_p$ and $N$ is the norm of $F/\mathbf Q$ (this is valid for any simple extension $\mathbf Q(\alpha)/\mathbf Q$). Writing $(X-1)f'(X)=pX^{p-1}$, we get immediately $D(F)=\pm p^{p-2}$, with $+$ holding iff $p\equiv 1$ mod $4$ (see e.g. D. Marcus, "Number Fields", chap.2, ex.8). Recalling that $D(F)$ is a square in $F$ and using the uniqueness of $K$, we get $K=F(\sqrt {p^*})$, where $p^*=(-1)^{\frac {p-1}2}p$ . In your example here, $K=\mathbf Q(\sqrt {-7})$.

NB: The ramification criterion used by @Lubin amounts to the computation of $D(F)$.