6

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.

non-abelian group of order 9
  • 881
  • Is Eisenstein's criterion off-limits too, as well as Galois theory? – Angina Seng Apr 17 '17 at 05:38
  • I know Eisenstein's criterion. That is no problem. – non-abelian group of order 9 Apr 17 '17 at 05:46
  • 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? – Jyrki Lahtonen Apr 17 '17 at 07:46
  • 1
    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. – Paramanand Singh Apr 17 '17 at 07:53
  • @ParamanandSingh How should I prove that? – non-abelian group of order 9 Apr 17 '17 at 08:58
  • It's a standard theorem od Dedekind that if positive integers $m, n$ are coprime to each other then $[\mathbb{Q} (\zeta_{m}, \zeta_{n}) :\mathbb{Q} (\zeta_{n})] =[\mathbb{Q} (\zeta_{m}) :\mathbb{Q}] =\phi(m) $. – Paramanand Singh Apr 17 '17 at 09:04
  • @ParamanandSingh Does this theorem have a name so that I can google it? – non-abelian group of order 9 Apr 17 '17 at 10:00
  • You need to study a bit about cyclotomic polynomials. One reference is my answer https://math.stackexchange.com/a/532977/72031 but it only states this theorem at the end and does not prove it. The proof is similar to the proof given in that answer but slightly more complicated. You can also search for irreducibility of cyclotomic polynomials over cyclotomic extensions. – Paramanand Singh Apr 17 '17 at 10:11

1 Answers1

2

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] $.

Paramanand Singh
  • 87,309