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I want to find a Galois extension $K/\mathbb{Q}$ such that $[K:\mathbb{Q}]=3$. I thought about this for a while, but haven't been able to come up with one yet.

What I tried so far: (i) Taking a separable polynomial $f\in\mathbb{Q}[x]$ of degree three and considering its splitting field. (ii) Looking at the splitting fields of primitive roots of unity.

The second one doesn't work because the splitting field over such a root has as degree a value in the range of Euler's totient function, and this doesn't contain three.

The first approach also didn't work. I tried polynomials of the form $(x-\sqrt{p})(x-\sqrt{q})(x-\sqrt{r})$ for primes, but those have degree $8$. I then tried 'third roots' $\alpha$, but the minimal polynomials of those have complex as well as real roots, so the simple extensions $K(\alpha)$ aren't normal unless they're trivial.

Could anyone please give me a hint on what else to try.

Adam Hughes
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azureai
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  • A useful discussion can be found here – lulu Jan 24 '17 at 20:21
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    My post here gives an easy-to-check necessary and sufficient condition on an irreducible cubic $f$ for $\text{Gal}(f/\mathbb{Q}) \cong \mathbb{Z}_3$ (warning: heavy spoilers). http://math.stackexchange.com/questions/2017810/galois-group-of-irreducible-cubic-equation/2017815#2017815 – Kaj Hansen Jan 24 '17 at 20:36

3 Answers3

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Consider the cyclotomic polynomial $\Phi_7(x) = x^6+x^5+\ldots + x + 1$ which is irreducible and generates an extension of $\Bbb Q$ of degree $6$ which is abelian (i.e. it is Galois with abelian Galois group). Then if $\zeta_7$ is a primitive $7^{th}$ root of $1$, $F=\Bbb Q(\zeta_7)$ is the extension. The element $\zeta_7+\zeta_7^{-1}$ is fixed by complex conjugation (an element of order $2$) and no other automorphism (you can check directly by noting $\zeta_7\mapsto \zeta_7^{k}, 1\le k\le 6$ are the automorphisms of $F$ and that any other automorphism besides $k=6$ gives a different element.

But then $K= \Bbb Q(\zeta_7+\zeta_7^{-1})\subseteq F$ is an extension of degree $3$, because that is the index of the fixing Galois group generated by complex conjugation. Hence $K/\Bbb Q$ is the desired extension. You can even describe it explicitly as $K=\Bbb Q\left((\cos\left({2\pi\over 7}\right)\right)$.

Working out the details you can see it is generated by the polynomial

$$p(x) = x^3+x^2-2x-1.$$

Adam Hughes
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  • @see you do not need all of the subgroups, you know that complex conjugation generates some subgroup of order two, and that implies the index is three. The other subgroups are not relevant for the purposes of this question. – Adam Hughes Jan 24 '17 at 23:04
  • @see No, not quite, $\zeta_7\mapsto \zeta_7^0$ is not an automorphism as it sends something not $1$ to $1$, you mean $1\le k\le 6$. Complex conjugation is clearly a field automorphism because $F\subseteq\Bbb C$ is a subfield. It is exactly the map $\zeta_7\mapsto\zeta_7^6=\zeta_7^{-1}$, but I chose not to bother looking at it from that point-of-view because why complicate matters when "complex conjugation" is a much more familiar way to phrase it. If it helps you to think of it the other way you can, they're the same thing, but remember, you don't need to know the full Galois group (contd) – Adam Hughes Jan 25 '17 at 00:46
  • @see all you need is its order and the order of complex conjugation, since you know by Lagrange's theorem that a subgroup of order two in a group of order six has index $3$, and the FTGT tells you that the extension degree of the fixed field is this index--i.e. $3$--which is all the problem asks for. – Adam Hughes Jan 25 '17 at 00:48
  • @see yes it does, but it's not relevant to this question. (I get why you want to know though). If you really are interested in it, complex conjugation clearly permutes the roots by noting

    $$\zeta_7^k\leftrightarrow \zeta_7^{-k}=\zeta_7^{7-k}.$$

    So if you number the roots $\alpha_i= \zeta_7^i, 1\le i\le 6$ then the permutation is just $(1, 6)(2, 5)(3, 4)$ in the cycle notation of the symmetric group.

     – Adam Hughes Jan 25 '17 at 01:30
  • I thought it is relevant for me because I learned the process to go this way: For each subgroup look at which basis elements stay fixed under the automorphisms in this group, then consider the subfield you get by adjoining them to the smaller field. This would be the subfield that corresponds to the subgroup. – azureai Jan 25 '17 at 01:40
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    @see yes, that's one approach, I opted for the faster route, since--as you'll find as you progress--reinventing the wheel every time takes a while, so we learn to use alternative ways with more general approaches. You will not want to work out entire Galois groups for every problem unless the problem asks you to find every subfield, and exploiting complex conjugation is one of the best tricks to add to your toolbox to avoid the excess work. – Adam Hughes Jan 25 '17 at 01:42
  • How do you come up with $\zeta_7+\zeta_7^{-1}$ then if you don't get it via the basis-vectors? This would be my last question and then I've understood it I guess. – azureai Jan 25 '17 at 01:44
  • @see it's a number plus its complex conjugate, so I know its fixed by complex conjugation since it's a real number just from basic facts about complex numbers (no basis required!) $x+iy+x-iy=2x\in\Bbb R$. – Adam Hughes Jan 25 '17 at 01:45
  • So just adjoining -any- element fixed under the subgroup-automorphisms will give me the proper subfield? – azureai Jan 25 '17 at 01:47
  • @see no, as I said in my answer it is not fixed by any other automorphism (I listed the others as well in the original answer). That is necessary. – Adam Hughes Jan 25 '17 at 01:48
  • Well we didn't cover this in class, we only did the basis-approach. So I need to read this up somewhere else. Thank you very much though. – azureai Jan 25 '17 at 01:49
  • @see I'm not sure what you mean by "the basis approach," can you clarify? – Adam Hughes Jan 25 '17 at 01:51
  • Given a finite Galois-extension $L|K$ of degree n, where L has a $K$-basis $1,\alpha,...,\alpha^{n-1}$, to determine the subfield corresponding to any subgroup $H$ of the Galois-group we would consider which elements of the above basis stay fixed under all $\sigma\in H$. Adjoining these to $K$ then gave the corresponding subfield. I tried to understand what you do from this perspective. Edit: All of the subgroups we considered where actually cyclic, and we only checked which basis-elements stay fixed under the generating automorphism. – azureai Jan 25 '17 at 01:57
  • @see Well if you want it from that point of view, we know from what I wrote earlier that $\zeta_7^k+\zeta_7^{-k}, 1\le k\le 3$ are fixed by the order-$2$ element complex conjugation given as an element of $S_6$ as $(1, 6)(2, 5)(3, 4)$. No element of the basis stays fixed except for $1$, and this is a good thing, because any power, $\zeta_7^j, 1\le j\le 5$ is another primitive $7^{th}$ root of $1$, and so fixing it would fix a generator of the field, and hence the whole field. What is fixed is combinations of basis vectors like those from earlier in the comment. In particular – Adam Hughes Jan 25 '17 at 02:01
  • @see $K=\Bbb Q(\zeta_7+\zeta_7^{-1}, \zeta_7^2+\zeta_7^{-2}, \zeta_7^3+\zeta_7^{-3})$ but a little computation shows $(\zeta_7+\zeta_7^{-1})^2 = 2 +\zeta_7^{-2}+\zeta_7^2$ so the second generator is not needed, and similarly $(\zeta_7+\zeta_7^{-1})^3=(\zeta_7^3+\zeta_7^{-3}) + 3(\zeta_y^2+\zeta_7^{-2})+6$ so the third generator is not needed. – Adam Hughes Jan 25 '17 at 02:03
  • But what is your argument then that $K(\zeta_7+\zeta_7^{-1})$ is the subfield corresponding to complex conjugation? It suffices to take any element that is fixed under one, but not any of the other automorphisms? – azureai Jan 25 '17 at 02:13
  • @see Recall that this is the definition of a fixed field of an element (or the subgroup generated by an element if we're being technically complete). It is that it is fixed by that element (i.e. the group generated by it). The key fact here is that a field is fixed iff its generator(s) is (are) fixed. – Adam Hughes Jan 25 '17 at 02:22
  • Our definition of fixed field is ${\alpha\in L|\sigma(\alpha)=\alpha \forall\sigma\in H}$ where $H$ is the subgroup. Adjoining any fixed element just gives me a subfield of this field. How do you know that it's the whole thing? – azureai Jan 25 '17 at 02:36
  • @see Clearly complex conjugation fixes the field $K$, so $K$ is contained in the fixed field. As no other automorphism fixes is $K$ is the whole field, this is the Fundamental Theorem of Galois theory. – Adam Hughes Jan 25 '17 at 02:40
  • Let us continue this discussion in chat. – azureai Jan 25 '17 at 02:59
  • @see First a small notational quibble: $K$ already has $\zeta_7+\zeta_7^{-1}$. Now, to the main issue: the FTGT tells you that $K$ is the fixed field, it's that theorem that gives us the result, it's not a tautology, we are using that theorem to say that $K$ is the fixed field, that fact is not self-evident. – Adam Hughes Jan 25 '17 at 03:00
  • I got it now. Thank you very much. – azureai Jan 25 '17 at 03:13
  • @see Glad I could help! – Adam Hughes Jan 25 '17 at 03:19
  • could you explain how you found the polynomial p? – marsarius Jul 31 '19 at 11:40
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    The minimal polynomial for any algebraic number is the product of linear terms in the conjugates of that number, form $(x-\alpha)$. Since you know the conjugates are all $\zeta^j+\zeta^{-j}$ it's fairly straightforward. – Adam Hughes Jul 31 '19 at 22:49
  • @marsarius There are some more details on that here. – Anakhand Mar 15 '22 at 14:24
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To get an Galois extension of degree $3$, you need all the roots to be real since otherwise complex conjugation is an automorphism of order $2$.

Whilst you're right that Euler's totient function doesn't take the value $3$, we can tweak this idea. Now let $\zeta_7$ be a primitive $7$th root of unity. Then $[\mathbb{Q}(\zeta_7):\mathbb{Q}]=6$ and it has cyclic Galois group. As I said above, complex conjugation creates an order 2 element so we shall take the fixed field corresponding to this.

Now the complex conjugate $\zeta_7$ is $\zeta_7^{-1}$ and it turns out that $[\mathbb{Q}(\zeta_7+\zeta_7^{-1}):\mathbb{Q}]=3$. Moreover, this extension is Galois since $\mathbb{Q}(\zeta_7)/\mathbb{Q}$ had an abelian Galois group.

In fact, all degree $3$ Galois extensions of $\mathbb{Q}$ will arise like this as being the subfield of some other cyclotomic field.

Matt B
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Hint: Consider the cyclotomic field $\mathbb{Q}(\zeta_7)$. where $\zeta_7=e^{2\pi i/7}$. Its Galois group is $\mathrm{Gal}(\mathbb{Q}(\zeta_7)/\mathbb{Q})\cong\mathbb{Z}/6\mathbb{Z}$, which is abelian and therefore all of its subgroups are normal. Can you think of a subgroup $H\subset \mathbb{Z}/6\mathbb{Z}$ such that the fixed subfield $\mathbb{Q}(\zeta_7)^H$ is a degree $3$ Galois extension over $\mathbb{Q}$? Use the Fundamental Theorem of Galois Theory.

Zev Chonoles
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  • You can more generally construct cyclic extensions of Q of degree n if you admit known properties of cyclotomic extensions of Q http://math.stackexchange.com/a/2028709/300700 – nguyen quang do Jan 25 '17 at 16:48