In the book Abstract Algebra of Dummit and Foote, there is a problem as follows :

Let $K=\mathbb{Q}(\sqrt[8]{2},i), F_1=\mathbb{Q}(i), F_2=\mathbb{Q}(\sqrt{2}), F_3=\mathbb{Q}(\sqrt{-2})$. Prove that: $Gal(K/F_1)\cong Z_{8}, Gal(K/F_2)\cong D_8, Gal(K/F_3)\cong Q_8$

Here is my argument :

Note that $K$ is the splitting field of $(x^8-2)(x^2+1)=0$ over $\mathbb{Q}$ so, $K$ is an extension field of $Q$ of degree 16. Since $F_1=\mathbb{Q}(i)$ is an extension field of degree 2 over $\mathbb{Q}$, using the fundamental theorem of Galois theory, we have |Gal(K/F_1)|=8$.

Now, my question is : How can I find the group $Gal(K/F_1)$ exactly and efficiently ? I have no way except trying to list all the possible elements, find the relationships between them, and then conclude, it may take too long for an extension of large degree.

  • $\begingroup$ I just noticed, there is no need to say $K$ is the splitting field for $(x^8-2)(x^2+1)$, as the latter factor is actually implicit in the former. I.e., $K$ is the splitting field of $x^8-2$. Of course, it's still true! But I figured it may be worth pointing out. $\endgroup$ – Alex Jan 6 '13 at 2:10

The field $K$ is an extension of $F_1$ of degree $8$. The minimal polynomial of $\sqrt[8]2$ over $F_1$ divides $x^8-2$ since $\sqrt[8]2$ is a root. Since adding $\sqrt[8]2$ to $F_1$ generates an extension of degree $8$ of $F_1$, the minimal polynomial of $\sqrt[8]2$ over $F_1$ must have degree $8$, hence it is $x^8-2$. Therefore the extension is cyclic, hence its Galois group is cyclic.

To check this last claim, use Proposition $36$ of chapter 14 section 7 : since $L = \Bbb Q(\sqrt 2, i)$ contains the $8^{\text{th}}$ roots of unity, the extension $L(\sqrt[8]2)/L = K/L$ is cyclic. Therefore, $\mathrm{Gal}(K/L)$ is a cyclic subgroup of $\mathrm{Gal}(K/F)$. Since $\mathrm{Gal}(K/L)$ has index $2$ in $\mathrm{Gal}(K/F)$, it is normal, so let $\sigma_1 \in \mathrm{Gal}(K/F) - \mathrm{Gal}(K/L)$. Check that $\sigma_1$ generates $\mathrm{Gal}(K/F)$ using the above information.

EDIT : I admit I can't finish the proof... I thought it had a neat end. I'll leave it there just in case someone can finish it.

Perhaps using chapter 14.7 is a little harsh because it comes later on, but if you look at the proof of that proposition you see that the ideas are not quite fancy, you can read them already and get a good idea of what it is about.

Hope that helps,

  • $\begingroup$ I am checking the last claim at the moment ; I'm not sure about it... give me a minute. $\endgroup$ – Patrick Da Silva Jan 4 '13 at 15:59
  • $\begingroup$ How did you deduce the extension is cyclic ? $\endgroup$ – Belgi Jan 4 '13 at 16:11
  • $\begingroup$ @Belgi : It wasn't trivial, so I added it. $\endgroup$ – Patrick Da Silva Jan 4 '13 at 16:11
  • $\begingroup$ Thanks for the added explanation! $\endgroup$ – Belgi Jan 4 '13 at 16:13
  • $\begingroup$ @Belgi : Wait... there's something wrong in my argument. I had the wrong generator for the Galois group of $K/F$. I still don't know if there's a neat way to show that any element outside the subgroup generates the Galois group of $K/F$ (Think of $Z_8$ and the elements $1,3,5,7$...) $\endgroup$ – Patrick Da Silva Jan 4 '13 at 16:15

One approach that works here is a result of Kummer. Suppose that $L = F(\sqrt[n]{a})$ for some $a\in F$. If $L/F$ is Galois, then $\operatorname{Gal}(L/F) \simeq \mathbb{Z}/d\mathbb{Z}$ for some $d\mid n$.

Edit: In general, the result is for extensions $L/F$ where $F$ contains an $n$-th root of unity. However, the assumption that $F$ contains an $n$-th root of unity is required only in showing that $L/F$ is Galois, which I believe you're already sure of from you question. Feel free to ask if I'm interpreting falsely!

In your case, you already know that $K/F_1$ is Galois, and so, if you know the Galois group has order 8, and must be $\mathbb{Z}/d\mathbb{Z}$ for some $d\mid 8$, you are done!

The proof of this result is fairly easy to push to full generality, so I'll just look at what's left for the case with $K/F_1$. As the roots of $x^8 - 2$ are of the form $\omega^i\sqrt[8]{2}$ for $1\leq i\leq 8$, where $\omega$ is a primitive 8-th root of unity, each automorphism must map $\sqrt[8]{2}$ to one of these multiples, which is well-defined mod 8. So you have a map $\operatorname{Gal}(K/F_1) \to \mathbb{Z}/8\mathbb{Z}$ that is (check!) an injective homomorphism, and so $\operatorname{Gal}(K/F_1)$ is a subgroup of $\mathbb{Z}/8\mathbb{Z}$.

I don't think there is a "general method" that easily computes the Galois group of any extension, ever, (I could see showing an extension has Galois group isomorphic to the Monster needing particularly resilient methods of attack!) but if you know tricks that handle most of the common extensions, you should be able to deal with most extensions that come your way.

  • $\begingroup$ I think this result of Kummer requires the assumption that $F$ contains the $n^{\text{th}}$ roots of unity, which is why I was having trouble in my answer. Can you put a reference to your result? $\endgroup$ – Patrick Da Silva Jan 5 '13 at 14:42
  • $\begingroup$ @PatrickDaSilva, Hm, perhaps I missed something? The result does indeed use that $F$ contain an $n$-th root of unity (will add shortly), but it is only used to show that $L/F$ is Galois (otherwise, $L/F$ may not even be Galois!) But in our case, we know that it is already (unless I'm misinterpreting the question?) In particular, the map $\sqrt[8]{2}\mapsto \omega\sqrt[8]{2}$ is an automorphism of degree 8. $\endgroup$ – Alex Jan 5 '13 at 18:30
  • $\begingroup$ The result I know is this one : If $F$ contains all $n^{\text{th}}$ roots of unity, then $F(\sqrt[n]a)/F$ is a cyclic Galois extension of degree dividing $n$. If $F$ does not contain all $n^{\text{th}}$ roots of unity, I don't know any general result. That's why I was asking for a reference for your result (or a proof) because I have never seen this before. I just read your edit ; I'll take a look at the proof I have and I'll confirm what you said. $\endgroup$ – Patrick Da Silva Jan 5 '13 at 20:36
  • $\begingroup$ @PatrickDaSilva, I could very well be making a silly mistake! For the problem above: In $K$, we have $$ x^8 - 2 = x^8 - (\sqrt[8]{2})^8) = \prod_{j=1}^8 (x - \omega^j\sqrt[8]{2}),$$ and as $x^8-2$ is irreducible, we know any element of $\operatorname{Gal}(K/\mathbb{Q})$ sends $\sqrt[8]{2}$ to another root of $x^8-2$, so the above argument gives a copy of $\mathbb{Z}/8\mathbb{Z}$ for the automorphisms here. As $i$ is not a root of $x^8-2$, these lift to automorphisms of $K/F_1$, and so if we know there should only be 8, we are done? $\endgroup$ – Alex Jan 6 '13 at 2:07
  • $\begingroup$ @PatrickDaSilva, as for the general case, here's what I know. If $F$ contains a primitive $n$-th root of unity, $\omega$, then $L := F(\sqrt[n]{a})$ (where $a\in F$) is the splitting field for $$ x^n - a = x^n - (\sqrt[n]{a})^n = \prod_{j=1}^n (x - \omega^j a),$$ as $\omega\in F$. Thus $L/F$ is Galois. Now, any $\sigma\in\operatorname{Gal}(L/F)$ sends $\sqrt[n]{a}$ to a root of $x^n - a$, and we proceed as above, as the map $\operatorname{Gal}(L/F)\to \mathbb{Z}/n\mathbb{Z}$ is a well-defined, injective homomorphism. $\endgroup$ – Alex Jan 6 '13 at 2:07

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