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Put $z_{n} = e^{2\pi i /n}$. I am searching for $n \in \mathbf{N}$ so that $\mathbf{Q}(z_{5},z_{7}) = \mathbf{Q}(z_{n})$.

I know that : $z_{5} = \cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5}) $ and $z_{7} =\cos(\frac{2\pi}{7})+i\sin(\frac{2\pi}{7})$.

Can you give me a hint how to continue my search? Thank you.

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Hint: $z_5$ and $z_7$ are roots of unity. Are they powers of a common root of unity? –  lhf Dec 22 '11 at 15:48
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Hint: What is the degree of $\mathbb{Q}(z_n)/\mathbb{Q}$, and the degree of $\mathbb{Q}(z_5,z_7)/\mathbb{Q}$? –  Álvaro Lozano-Robledo Dec 22 '11 at 15:56
    
The hint about degrees may not be so helpful, as there can be many $n$ for which ${\bf Q}(z_n)$ has the same degree. –  Gerry Myerson Dec 22 '11 at 17:43
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Another hint: contemplate $z_5z_7$. –  Gerry Myerson Dec 22 '11 at 17:43
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@Tashi: No! $12\cdot12=144\equiv4\pmod{35}$. I was moving the problem from the product of complex powers of $e$ to the addition of their exponents. IOW I applied (the inverse of) the homomorphism $f:\mathbf{Z}/35\mathbf{Z}\rightarrow \mathbf{C}$ that maps $n+35\mathbf{Z}\mapsto e^{2\pi i n/35}$. The group operation in the residue class group is addition. –  Jyrki Lahtonen Dec 22 '11 at 20:16
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1 Answer

That's one too many hints in the comments, but the OP still seems in doubt, so here is a proof that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})$, where $\zeta_n=e^{2\pi i/n}$ is a primitive $n$th root of unity.

First, let us show that $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$. Notice that $$\zeta_{35}^7=(e^{2\pi i/35})^7 = e^{2\pi i/5}=\zeta_5.$$ Thus, $\zeta_5\in \mathbb{Q}(\zeta_{35})$. Similarly, $\zeta_7 = \zeta_{35}^5 \in \mathbb{Q}(\zeta_{35})$. Hence, $\mathbb{Q}(\zeta_5,\zeta_7)\subseteq\mathbb{Q}(\zeta_{35})$.

Next, let us show that $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Indeed, consider $$(\zeta_5\cdot\zeta_7)^3 = (e^{2\pi i/5}\cdot e^{2\pi i/7})^3 = (e^{2\pi i\cdot 12/35})^3 = e^{2\pi i\cdot 36/35} = e^{2\pi i}\cdot e^{2\pi i/35} = 1 \cdot e^{2\pi i/35} = \zeta_{35}.$$ Thus, $\zeta_{35}=(\zeta_5\cdot\zeta_7)^3\in \mathbb{Q}(\zeta_5,\zeta_7)$, and this shows the inclusion $\mathbb{Q}(\zeta_{35})\subseteq \mathbb{Q}(\zeta_5,\zeta_7)$. Therefore, we must have an equality of fields.

Now, suppose that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_n)$ for some $n\geq 1$. We have just shown that $n=35$ works. Are there any other possible values of $n$ that work? Well, if $n$ is odd, then $\mathbb{Q}(\zeta_n) = \mathbb{Q}(\zeta_{2n})$, so $n=70$ also works.

Finally, one can show that if $\mathbb{Q}(\zeta_m)\subseteq \mathbb{Q}(\zeta_n)$, then $m$ divides $n$ (here neither $m$ or $n$ should be twice an odd number). In particular, since we know that $\mathbb{Q}(\zeta_5,\zeta_7)=\mathbb{Q}(\zeta_{35})=\mathbb{Q}(\zeta_n)$, then $n$ is divisible by $35$, and therefore $\varphi(n)$ is divisible by $24$. If $n>70$ and divisible by $35$, then $\varphi(n)$ would be strictly larger than $24$, and that would be a contradiction, because $\varphi(n)$ is the degree of the extension $\mathbb{Q}(\zeta_n)/\mathbb{Q}$. Hence, $n=35$ or $70$.

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