Let $n\in\Bbb Z_{>0}$. Determine the Galois group of $f(x)=x^n+1$ over $\Bbb Q$.

I am having some trouble with this. I started by assuming $n$ is odd, then $f(-x)=(-x)^n+1=-(x^n-1)$, then the Galois group of $f(x)$ is the same as $x^n-1$. We know that $\operatorname{Gal}(x^n-1/\Bbb Q)\cong (\Bbb Z/n\Bbb Z)^\times$, so this is the Galois group of $f(x)$ over $\Bbb Q$ for odd $n$.

I am not sure what to do for the general case ($n$ odd or even). I have been doing some research, and I have seen people argue that the splitting field for $x^n+1$ over $\Bbb Q$ is equal to the one for $x^{2n}-1$ over $\Bbb Q$, since $x^{2n}-1=(x^n+1)(x^n-1)$, so any solution to $x^n+1=0$ is one to $x^{2n}-1=0$. For instance, the accepted answer here. Then I would be able to conclude that $\operatorname{Gal}(x^n+1/\Bbb Q)\cong(\Bbb Z/2n\Bbb Z)^\times$. However, I do not understand why this allows us to conclude that the splitting field for $x^n+1$ is $\Bbb Q(\zeta_{2n})$ though ($\zeta_{2n}$ a primitive $2n$-th root of unity).


All roots of $x^n-1$ have order dividing $n$, and there is a root of $x^{2n}-1$ of exact order $2n$. Since $$x^{2n}-1 = (x^n+1)(x^n-1)$$ we know there's a root of $x^n+1$ of exact order $2n$, say $\zeta$. The roots of $x^{2n}-1$ are precisely $$1,\zeta,\dots,\zeta^{2n-1}$$ So the roots of $x^n+1$ are a subset of this containing $\zeta$. So the splitting field of $x^n+1$ is $\mathbb{Q}(\zeta)$, which is the same as the splitting field of $x^{2n}-1$. We conclude that the Galois group is isomorphic to $(\mathbb{Z}/2n\mathbb{Z})^{\times}$.

Note that it's not necessary to consider two seperate cases: if $n$ is odd, then $$(\mathbb{Z}/2n\mathbb{Z})^{\times} \cong (\mathbb{Z}/2\mathbb{Z})^{\times}\times (\mathbb{Z}/n\mathbb{Z})^{\times} \cong (\mathbb{Z}/n\mathbb{Z})^{\times}$$ So indeed, we get the same as what you already noticed in the beginning of your question.

  • $\begingroup$ Just to clarify, by "order" you mean multiplicative order in the multiplicative group generated by the primitive roots of unity? So we can argue that $x^n+1$ has an order $2n$ root $\zeta$, so when we take powers of $\zeta$ we obtain all roots of $x^{2n}-1$, so $\Bbb Q(\zeta)=\Bbb Q(\zeta_{2n})$? $\endgroup$ – Dave Mar 8 '18 at 21:46
  • $\begingroup$ yes and yes on both questions. $\endgroup$ – Jef L Mar 8 '18 at 21:49
  • $\begingroup$ Okay, this really clears it up for me. Thank you! $\endgroup$ – Dave Mar 8 '18 at 21:50

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