I am trying to find an example of degree-100 extension of $\Bbb Q(\zeta_5)$ and an example of degree-100 extension of $\Bbb Q(\sqrt[3]{2})$.

For the example of degree-100 extension of $\Bbb Q(\sqrt[3]{2}),$ I suspect that $\Bbb Q(\sqrt[3]{2},\sqrt[100]{-3})$ is an example. Since $3$ and $100$ are co-prime, that means I can prove $[\Bbb Q(\sqrt[3]{2},\sqrt[100]{-3}):\Bbb Q]$ is at least $300$ and at most $300.$ Therefore $[\Bbb Q(\sqrt[3]{2},\sqrt[100]{-3}):\Bbb Q]=300.$ Then by the tower law, problem solved.

For the example of degree-100 extension of $\Bbb Q(\zeta_5)$, I really don't know that how to find (and prove) such an example.

Note: I haven't learnt Galois Theory. So please don't use that.Thanks so much.

  • $\begingroup$ Is Eisenstein's criterion off-limits too, as well as Galois theory? $\endgroup$ Apr 17, 2017 at 5:38
  • $\begingroup$ I know Eisenstein's criterion. That is no problem. $\endgroup$ Apr 17, 2017 at 5:46
  • $\begingroup$ Have you covered cyclotomic polynomials? For example, do you know the extension degrees of $\Bbb{Q}(\zeta_{500})$,$\Bbb{Q}(\zeta_{1000})$, $\Bbb{Q}(\zeta_{2000})$ and such? $\endgroup$ Apr 17, 2017 at 7:46
  • 1
    $\begingroup$ Note that $101$ is prime and hence we can choose $\zeta_{101}=e^{2\pi i/101}$ so that you have $[\mathbb{Q} (\zeta_{101}):\mathbb{Q}]=100$. Also it can be proved that $[\mathbb{Q} (\zeta_{101},\zeta_{5}):\mathbb{Q}(\zeta_{5})]=100$ so you get desired extension. $\endgroup$
    – Paramanand Singh
    Apr 17, 2017 at 7:53
  • $\begingroup$ @ParamanandSingh How should I prove that? $\endgroup$ Apr 17, 2017 at 8:58

1 Answer 1


Based on my comments I give the following answer. It uses only the irreducibility of cyclotomic polynomials in $\mathbb{Q}[x]$ and avoids the theorem of Dedekind mentioned in comments.

Let $\zeta_{n} = e^{2\pi i/n}$. Then $\zeta_{n}$ is a primitive $n$th root of unity there are $\phi(n)$ such primitive $n$th roots of unity given by $\zeta_{n}^{r}$ where $1 \leq r \leq n$ is and $r$ is coprime to $n$. The polynomial $$\Phi_{n}(x) = \prod_{1 \leq r \leq n, (r, n) = 1}(x - \zeta_{n}^{r})$$ is having integer coefficients and is irreducible in $\mathbb{Q}[x]$ so that $[\mathbb{Q}(\zeta_{n}):\mathbb{Q}] = \phi(n)$.

Next let $m, n$ be positive integers coprime to each other. Then we have integers $a, b$ such that $am + bn = 1$ and therefore $$\zeta_{mn} = \zeta_{n}^{a}\zeta_{m}^{b}$$ and clearly $$\zeta_{m} = \zeta_{mn}^{n}, \zeta_{n} = \zeta_{mn}^{m}$$ so that $$\mathbb{Q}(\zeta_{mn}) = \mathbb{Q}(\zeta_{m}, \zeta_{n})$$ And then $$[\mathbb{Q}(\zeta_{mn}): \mathbb{Q}] = \phi(mn) = \phi(m)\phi(n)$$ Let's put $m = 5, n = 101$ so that $\phi(m) = 4, \phi(n) = 100, \phi(mn) = 400$. We now have $$\mathbb{Q} \subset \mathbb{Q}(\zeta_{m})\subset\mathbb{Q}(\zeta_{mn})$$ and $$[\mathbb{Q}(\zeta_{mn}):\mathbb{Q}(\zeta_{m})] = \frac{[\mathbb{Q}(\zeta_{mn}):\mathbb{Q}]}{[\mathbb{Q}(\zeta_{m}):\mathbb{Q}]} = \frac{\phi(mn)}{\phi(m)} = \phi(n)$$ so that $\mathbb{Q}(\zeta_{mn}) = \mathbb{Q}(\zeta_{505})$ is our desired field extension.

From the above argument we see that if $m, n$ are coprime to each other then $$\mathbb{Q} (\zeta_{mn}) = \mathbb{Q}(\zeta_{m},\zeta_{n})=\mathbb{Q} (\zeta_{n}) (\zeta_{m}) $$ is a field extension of $\mathbb{Q} (\zeta_{n}) $ of degree $\phi(m) $. Moreover $\zeta_{m} $ satisfies a polynomial $\Phi_{m} (x) \in \mathbb{Q} [x] \subset\mathbb{Q} (\zeta_{n}) [x] $ of degree $\phi(m) $. It follows that the polynomial $\Phi_{m} (x) $ is irreducible in $\mathbb{Q} (\zeta_{n}) [x] $. Thus starting from the irreducibility of $\Phi_{n} (x) $ in $\mathbb{Q} [x] $ and using the theorem about degrees of a tower of field extensions we have proved the theorem of Dedekind referred in the beginning of the post:

Theorem: If $m, n$ are positive integers coprime to each other then the cyclotomic polynomial $\Phi_{m} (x) $ is irreducible in $\mathbb{Q} (\zeta_{n}) [x] $.

  • $\begingroup$ NIce and simple! Better than what I had in mind. $\endgroup$ Apr 17, 2017 at 21:27

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