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I have to show that the following statement is true:

Let $K$ Be a field extending $\mathbb{Q}$ such that $[K: \mathbb{Q}] \ = \ 2$. Then there is a root of unity $\zeta$ such that $K \subseteq \mathbb{Q}(\zeta)$.

What I did

Since $2$ is prime, there can't be a field $L$ such that $L \supset K \supset \mathbb{Q}$. That means that $ K \ = \ K(\alpha) \ \cong \ \mathbb{Q}[X] /(X^2 +aX +b) $ for some $a,b \in \mathbb{Q}$. Now I have to show that one always can find a number $n \in \mathbb{N}$, and $a_0, a_1, \cdots , a_n \in \mathbb{Q}$ such that $$ \alpha = a_0 + a_1\zeta + a_2\zeta^2 + \cdots + a_n-1\zeta^{n-1} $$ But I don't really know how.

Try to answer my own question using the hint

Since the roots of $X^2 +aX+b$ are $\frac{\pm\sqrt{a^2-4b}-a}{2}$, we only have to add $\sqrt{a^2-4b}$ to get the field $K$. Now we replace $m = a^2-4b$, and we split $m$ in prime numbers. $$ m \ = \ p_1^{k_1} \cdots p_t^{k_n} \quad \text{ so } \quad \sqrt{m} \ = \ \sqrt{p_1}^{k_1} \cdots \sqrt{p_t}^{k_n} $$ Now I write $\gamma_p \ := \ \sum_{a=0}^{p-1} \left( \frac{a}{p}\right) \zeta_p^a $. We know that $ \gamma_p^2 \ = \ p \left( \frac{-1}{p}\right) $, thus $ \gamma_p \ = \ \pm \sqrt{\pm p} $. Now we see that $$ \sqrt{m} \ = \ \pm g_{p_1}^{k_1}\cdots g_{p_t}^{k_t} \ \in \ \mathbb{Q}(\zeta_{p_1}, \cdots \zeta_{p_t}) $$ We know that $\zeta_{p_1} \cdot \zeta_{p_2} \cdots \zeta_{p_t}$ is a primitive root of unity as well. If we link up all this knowledge we have: $$ \mathbb{Q}(\alpha) \ = \ \mathbb{Q}(\sqrt{m}) \ = \ \mathbb{Q}(\pm g_{p_1}^{k_1}\cdots g_{p_t}^{k_t} ) \ = \ \mathbb{Q}(\zeta_{p_1}, \cdots \zeta_{p_t}) \ = \ \mathbb{Q}(\zeta_{p_1} \cdot \zeta_{p_2} \cdots \zeta_{p_t}) $$ Which means that the product of those primitive roots is the root of unity we had to find.

Jyrki Lahtonen
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Koenraad van Duin
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  • Here's a hint to step in a slightly different direction: can you show that you can take $\alpha$ to be $\sqrt{N}$ for some $N\in\Bbb{Z}$? – Steven Stadnicki Feb 21 '15 at 23:09
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    Do you mean $Q \subset L \subset K$? – Z. L. Feb 21 '15 at 23:09
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    For future reference: this is the first case of a very general, beautiful, and difficult theorem that says that if the Galois group of $K$ over the rationals is abelian, then $K$ is contained in a cyclotomic extension of the rationals. – Gerry Myerson Feb 21 '15 at 23:34
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    And that is the Kronecker–Weber theorem. – kahen Feb 21 '15 at 23:36
  • Related: https://math.stackexchange.com/questions/475370, https://math.stackexchange.com/questions/2004676, https://math.stackexchange.com/questions/282757, https://math.stackexchange.com/questions/1616900, https://math.stackexchange.com/questions/31485 – Watson Jan 31 '17 at 11:16

1 Answers1

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Hint: As said in the comments, every quadratic extension of $\mathbb{Q}$ is of the form $\mathbb{Q}(\sqrt{\alpha})$ with $\alpha\in\mathbb{Z}$. Try proving the assertion first for $K=\mathbb{Q}(\sqrt{\pm p})$ with $p$ prime. For that, consider the Gauss sum $\gamma = \displaystyle\sum_{a=0}^{p-1} \left(\dfrac{a}{p}\right)\zeta_p^a$ where $\zeta_p$ is a primitive $p$-th root of unity and $\left(\dfrac{a}{p}\right)$ is the Legendre symbol. Prove that $\gamma^2 = p \left(\dfrac{-1}{p}\right)$. Prove that $\gamma$ is fixed by the only subgroup of $\mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ of index $2$.

Nacho Darago
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  • One can also work with the sum $\sum_{a=0}^{p-1}\zeta_p^a$. – Gerry Myerson Feb 21 '15 at 23:36
  • That sum is zero. Maybe you meant something else? – Nacho Darago Feb 21 '15 at 23:48
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    Sorry, $\sum_{a=0}^{p-1}\zeta_p^{a^2}$. – Gerry Myerson Feb 21 '15 at 23:49
  • @Nacho Darago, thank you for your hint! I have tried to solve the assignment now, I hope that you can check whether I understood everything. – Koenraad van Duin Feb 22 '15 at 09:10
  • Seems right. When you say $\mathbb{Q}(\pm g_{p_1}^{k_1}\ldots g_{p_t}^{k_t}) = Q(\zeta_{p_1},\ldots , \zeta_{p_t})$ you mean $\mathbb{Q}(\pm g_{p_1}^{k_1}\ldots g_{p_t}^{k_t}) \subseteq Q(\zeta_{p_1},\ldots , \zeta_{p_t})$. Also, you should still prove that $\gamma_p^2 = \left(\dfrac{-1}{p}\right) p$. However, all the ideas seem fine. – Nacho Darago Feb 22 '15 at 22:10