The splitting field of $x^{20}-1$ over $\mathbb Q$

Question

I am trying to find the splitting field of $x^{20}-1$ over $\mathbb Q$. I know that it has degree $8$ over $\mathbb Q$, and that $i$ and $\sqrt5$ are in the splitting field. I am also suspecting that $\sqrt[4]5$ is in the splitting field, which would mean that the splitting field is $\mathbb Q(i, \sqrt[4]5)$, but am uncertain about how to prove this. Could anyone help me on those regards?

EDIT: After consideration, the splitting field is most likely $\mathbb Q(i, \sqrt{10+2\sqrt5})$, but how does one show this without the exact trigonometric value?

Answer 1

Let $S$ be the splitting field of $x^{20}-1$ over $\Bbb Q$.

Let $i\not=\pm1$ be a root of $x^4-1=0$, i.e., $i$ is the imaginary unit.

Let $\alpha\not=1$ be a root of $x^5-1=0$.

Consider $i^j\alpha^k$, where $0\le j<4$ and $0\le k<5$.

• $(i^j\alpha^k)^{20}=(i^{20})^j(\alpha^{20})^k=1\cdot1=1.$

• Suppose $i^{j_1}\alpha^{k_1}=i^{j_2}\alpha^{k_2}$, where $0\le j_1,j_2<4$, $0\le k_1,k_2<5$. Then $i^{j_2-j_1}\alpha^{k_2-k_1}=1$.

  • $1=(i^{j_2-j_1}\alpha^{k_2-k_1})^4=\alpha^{4(k_2-k_1)}=\frac1{\alpha^{k_2-k_1}}$. Hence $k_2=k_1$.
  • $1=(i^{j_2-j_1}\alpha^{k_2-k_1})^5=i^{5(j_2-j_1)}=i^{j_2-j_1}$. Hence $j_2=j_1$.

Hence, all roots of $x^{20}-1$ are $i^j\alpha^k$. The splitting field is $Q(i,\alpha)$.

(More generally, the splitting field of $x^{ab}-1$ is the product field of the splitting field of $x^a-1$ and that of $x^b-1$ when $\gcd(a,b)=1$).

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What are the roots of $x^5-1=0$?

Since $x^5-1=(x-1)(x^2+\frac{1+\sqrt5}2x+1)(x^2+\frac{1-\sqrt5}2x+1)$, the roots are
$(1, \frac{-1- \sqrt 5\pm i\sqrt{10 - 2\sqrt 5}}4, \frac{-1+ \sqrt 5\pm i\sqrt{10 + 2\sqrt 5}}4).$

Consider root $\alpha = \frac{-1+ \sqrt 5+ i\sqrt{10 + 2\sqrt 5}}4\not=1$. We know $$S=\Bbb Q(i,\alpha),$$
which can be the sought solution.

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$\frac1\alpha=\frac{-1+ \sqrt 5- i\sqrt{10 + 2\sqrt 5}}4$.
$\sqrt 5=4(\alpha +\frac1\alpha)+\frac12$.
Hence $\sqrt5\in S$. So is $\sqrt{10+2\sqrt5}=(4\alpha+1-\sqrt5)/i$. Hence, $\Bbb Q(i, \sqrt{10+2\sqrt5})\subseteq S$.

One the other hand, since $\sqrt5=\frac{\left(\sqrt{10+2\sqrt5}\right)^2-10}2$, we have $\alpha\in \Bbb Q(i, \sqrt{10+2\sqrt5})$. Hence $S\subseteq \Bbb Q(i, \sqrt{10+2\sqrt5})$.

So we know $$ S=\Bbb Q(i, \sqrt{10+2\sqrt5}),$$ which can be the sought solution, too, as you suspected.

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Similarly, we can obtain $$S=\Bbb Q(i, \beta)=\Bbb Q(i, \sqrt{10-2\sqrt 5})$$ using $\beta=\frac{-1- \sqrt 5+ i\sqrt{10 - 2\sqrt 5}}4$, another root of $x^5=1$.

It must be true that $\sqrt{10 - 2\sqrt 5}\in Q(i, \sqrt{10+2\sqrt 5})$. Here is how we can see this containment directly. Note that $(10-2\sqrt5)(10+2\sqrt5)=80$, we have $$\sqrt{10-2\sqrt5}=\frac{4\sqrt5}{\sqrt{10 + 2\sqrt 5}}\in Q(\sqrt{10 + 2\sqrt 5}).$$

Answer 2

Hint

Take a primitive $20$th root of unity $\zeta$. Its minimal polynomial is $\Phi_{20}$, the twentieth cyclotomic polynomial over $\Bbb Q.$ It's irreducible by a result of Gauss's.

Then you want $\Bbb Q(\zeta).$

Since $\cos(\pi /{10})=1/4(\sqrt{2\sqrt5 +10}),$ we should have $\Bbb Q(\zeta) =\Bbb Q(i,\sqrt{2\sqrt5 +10}).$