6
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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

2
$\begingroup$

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] $.

$\endgroup$
1
  • $\begingroup$ NIce and simple! Better than what I had in mind. $\endgroup$ Apr 17, 2017 at 21:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.