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Prove that the $7$-th cyclotomic extension $\mathbb{Q}(\zeta_7)$ contains $\sqrt{-7}$

I thought that the definition of the $n$-th cyclotomic extension was: $\mathbb{Q}(\zeta_n)=\{\mathbb{Q}, \sqrt{-n}\}$. Is this correct?

How could I prove the statement? Could we consider the polynomial $X^2+7$ (which is irreducible by Eisenstein's criterion with $p=7$?

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    $\begingroup$ @DietrichBurde The question to which you link subsumes the case here, but there is a somewhat simpler proof in this particular case (as I've provided below)... $\endgroup$ – Benjamin Dickman May 31 '16 at 18:18
  • $\begingroup$ @BenjaminDickman Yes, there is a simpler proof. Like having a proof that the sum of the first odd numbers is a square, and you provide an easier proof that $1+3+5+7=16$ is a square. Still I think, the linked duplicate is the right way to consider this question. Also, there was the confusion of the OP about adjoining $\sqrt{-n}$ instead of $\zeta_n$. $\endgroup$ – Dietrich Burde May 31 '16 at 18:21
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    $\begingroup$ @DietrichBurde Perhaps for some; however, the top answer to which you link involves recalling that "only $p$ can ramify in $\mathbb{Q}(\zeta_p)$" ... the other answers involve a quadratic Gauss sum, and the Vandermonde determinant. I guess from the OP's question here around Eisenstein's criterion as an approach that the more general case involves un-encountered terms. That is, the "right way to consider the question" is a function of the considerer... $\endgroup$ – Benjamin Dickman May 31 '16 at 18:27
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    $\begingroup$ Yes, I agree that one can take this point of view. The other answers here, however, are not too difficult, and still are in the direction of the duplicate. So, I think, it is a matter of taste. $\endgroup$ – Dietrich Burde May 31 '16 at 18:30
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Let $\alpha = \zeta_7 + \zeta_7^{2} + \zeta_7^{4} \in \mathbb{Q}(\zeta_7)$.

What is $\alpha$? Well, observe $\alpha^2 + \alpha + 2 = 0$.

By the quadratic equation, we find:

$$\alpha = \frac{-1 \pm \sqrt{-7}}{2}$$

whence $\sqrt{-7} \in \mathbb{Q}(\zeta_7)$ as desired. QED.

(In particular, the $\pm$ should be $+$ although either case yields $\sqrt{-7} \in \mathbb{Q}(\zeta_7)$ as desired.)

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    $\begingroup$ (cf. Example 3.0.8 in paul garrett's notes.) $\endgroup$ – Benjamin Dickman May 31 '16 at 18:20
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    $\begingroup$ Good. Efficient and elegant. $\endgroup$ – Lubin Jun 1 '16 at 2:45
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I'm sure there are more elegant ways to observe this, but here's one possible way:

  1. First, show that the extension is Galois (not too hard) and that $Gal(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong Z_6$. This has subgroups of order 2 and 3.

  2. Now find the fixed field of the subgroup of order 3. This is a degree 2 extension. To find the generator, find a generator of the Galois group and write the subgroup in terms of powers of the generator, and find a generator of the subgroup. Look at the orbit of the generator of that subgroup on $\zeta_7$. We guess that this will generate a degree 2 extension.

  3. To show that, take powers of that generator and try to find the minimal polynomial. It should be degree 2. The discriminant of the quadratic will be $\sqrt{-7}$, and you will be done.

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The cyclotomic field $\mathbb{Q}(\zeta_n)$ is defined by adjoining a primitive $n$-th root of unity, and we have $[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\phi(n)$. In particular, it is different from $\mathbb{Q}(\sqrt{-n})$ for $n>3$. However, every cyclotomic field $\mathbb{Q}(\zeta_p)$ for $p$ prime contains a unique quadratic subfield $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, or respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 \pmod{4}$, see here. Since $7\equiv 3 \bmod 4$ the claim follows, i.e., we have $\mathbb{Q}(\sqrt{-7})\subseteq \mathbb{Q}(\zeta_7)$.

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Note that $\Bbb Q(\zeta_7)$ is a totally ramified extension above $(7)$ of $\Bbb Z$ with discriminant a power of $7$. But then all subextensions must be totally ramified at $(7)$ and nowhere else. This means their discriminants must divide $7^n$. Because $\phi(7)=6$ is even and the Galois group is cyclic, we know there is a unique quadratic subfield, all of which are of the form $\Bbb Q(\sqrt{D})$ for some $D$ square free, and we know the discriminant of this field is

$$\begin{cases} D & D\equiv 1\mod 4 \\ 4D & D\equiv 2,3\mod 4\end{cases}$$

So our $D$ must be $\pm 7$, but $+7\equiv 3\mod 4$ would be ramified at $2$, hence it must be that $D=-7\equiv 1\mod 4$.


Addendum The definition of the $n^{th}$ cyclotomic field is the splitting field of $x^n-1$, your definition would be the splitting field of $x^2+n$.

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  • $\begingroup$ Do you have a reference for "then all subextensions must be totally ramified at $(7)$ and nowhere else." ? I'm not sure to know why this is true. $\endgroup$ – Watson Jun 27 '16 at 20:37
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    $\begingroup$ @Watson this is standard ramification theory, see for example Marcus' Number Fields or anything about "decomposition," "inertia," and "ramification" fields of an arbitrary extension, then restrict to the special case of a totally ramified extension. $\endgroup$ – Adam Hughes Jun 28 '16 at 0:54
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The method for showing that CZ7 (the cyclotomic numbers on the heptagon, satisfying $x^7+1=0$ and the span of powers of x) contains the $\sqrt{-7}$ is to realise that it is a sixth-order system, and one can dissect an order-two unit by merging the order-3 units out of it, for example, cis2/14 + cis 4/14 + cis 8/14 is invariant to the heptagonal isomorphism, and contains a form in the type x+y sqrt(-7), where these are integers or integer-halfs.

Z7 represents the span of chords of the heptagon. CZ7 represents the cyclotomic numbers form by dividing the half-circle into seven parts, and taking the span over Z. This represents the verticies of the tiling {7, 14/5}. It has a construction in terms of a general carryless base, modulo by setting a polynomial to zero, as $Pm(x^7+1=0)$. $P$ here represents $\sum z_j x^j$, where z,j are in the integers Z and $m(\mbox{eqn})$ represents an equity equation.

Some elementry number theory can prove that if $p$ is a prime of the form $4x+3$, then the cyclotomic numbers CZp contain $Z(1, \frac 12(1+\sqrt{-p}))$.

The rule of isomorphism says that there are several solutions to this equation, and putting eg x + x' + x" = y gives a value automorphic to this value. In the case of CZ7, an automorphic solution can be had by replacing $x$ by $x^9$ by $x^{11}$ by $x$. So the triplets $x+x^9+x^{11}$, and $x^2+x^4+x^8$ etc, are automorphic to the $x$ -> $x^9$ transform.

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  • $\begingroup$ I've added more to the answer. $\endgroup$ – wendy.krieger Jun 2 '16 at 9:11

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