For $n\in \mathbb N$, the splitting field of $x^n-2$ over $\mathbb Q$ has degree $n\cdot\phi(n)$.

Question

I suspect the following is true:

For $n\in \mathbb N$, the splitting field of $x^n-2$ over $\mathbb Q$ has degree $n\cdot\phi(n)$.

Clearly true when $n$ is relatively prime with $\phi(n)$, for example when $n$ is a prime. In the general case it will suffices to show that $x^n-2$ is irreducible over $\mathbb Q(\zeta_n)$. Since $\mathbb Q(\zeta_n)/\mathbb Q$ is a(Galois) normal extension all irreducible factors of $x^n-2$ in $\mathbb Q(\zeta_n)[x]$ has equal degree. But I can't show that $x^n-2$ is irreducible over $\mathbb Q(\zeta_n)$. I need some help. Thanks.

More about the case $n=8$. Sorry about misleading you a bit in your previous question. Anyway, the posted answer (+1) gets to the point. — Jyrki Lahtonen, Jun 29 '21 at 12:01
More generally the square root of any integer belongs to some cyclotomic field. A deep result know as Kronecker-Weber theorem tells us that every abelian Galois extension of $\Bbb{Q}$ is contained in some cyclotomic field. The quadratic extensions have a cyclic Galois group of order two, so... :-) — Jyrki Lahtonen, Jun 29 '21 at 12:06
But, by that argument from the earlier question no odd order roots of primes belong to cyclotomic extensions, so here only even values of $n$ need more care. — Jyrki Lahtonen, Jun 29 '21 at 12:08
Does this answer your question? Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$ — Jianing Song, May 01 '24 at 02:23

score 3 · Accepted Answer · answered Jun 29 '21 at 11:51

3

Counterexample: $\sqrt2\in\mathbb{Q}(\zeta_8)=\mathbb{Q}(\sqrt{2},\sqrt{-1})$ so the splitting field of $X^8-2$ is $\mathbb{Q}(\sqrt[8]2,\sqrt{-1})$ with degree $16$ over $\mathbb{Q}$, not $8\cdot\phi(8)=32$.

answered Jun 29 '21 at 11:51

user10354138

33,887

For $n\in \mathbb N$, the splitting field of $x^n-2$ over $\mathbb Q$ has degree $n\cdot\phi(n)$.

1 Answers1