# If $a = \sqrt{1+\sqrt{2}}$ and $E = \mathbb{Q}(a)$, find the minimal polynomial of $a$ over $\mathbb{Q}$, $[E:\mathbb{Q}]$, and $Gal(E / \mathbb{Q})$.

I have found the minimal polynomial for this problem to be $m(x) = x^4 -2x^2 -1$. So then $[E:\mathbb{Q}]$ is $4$ because that's the degree of the minimal polynomial. The next step (finding the Galois group) is where I'm struggling. My first instinct was to look at the roots and what they would map to, those being:

$\sqrt{1+\sqrt{2}} \rightarrow \sqrt{1+\sqrt{2}}$ and $\sqrt{1+\sqrt{2}} \rightarrow -\sqrt{1+\sqrt{2}}$

So I thought that the Galois group would have order $2$ leading me to believe it was $\mathbb{Z}_2$, however I remembered that $|Gal(E/\mathbb{Q})| = [E:\mathbb{Q}]$. So I knew something was wrong. Any tips on how to go about this problem from here?

• Sub-extensions correspond to subgroups of the Galois group. The two groups of order $4$ has different number of subgroups. How many sub-extensions are there? Maybe you could find two, or prove there is only one? – Arthur Jul 26 '17 at 17:59

Sorry about my earlier mistake (in a deleted comment), but on revisiting this page I noticed that there's something worth adding. Your question asks for the Galois group for $E : \mathbb Q$, not the Galois group for the splitting field of $x^4 - 2x^2 - 1$ over $\mathbb Q$. Since the extension $E : \mathbb Q$ is not normal, the statement $|{\rm Gal}(E : \mathbb Q)| = [E: \mathbb Q]$ is false, and your claim that $|{\rm Gal} (E : \mathbb Q) | = 2$ is correct!.

I'll explain in more detail, but first, we should agree on some notation: $$a = \sqrt{1 + \sqrt{2}}, \ \ \ \ \ \ b = \sqrt{1 - \sqrt{2}}.$$

As Lord Shark explained, the splitting field of $x^4 - 2x^2 - 1$ is not $E$. It is actually $E(b)$, which is strictly larger than $E$ since $b$ is imaginary whereas everything in $E$ is real. Therefore, $E : \mathbb Q$ is not a normal extension.

[One characterisation of what it means for an extension $L : K$ to be normal is that every irreducible polynomial $p(x) \in K[x]$ with one root in $L$ splits completely over $L$. Here, the polynomial $x^4 - 2x^2 - 1$ is irreducible in $\mathbb Q[x]$, but it has two roots $\pm a$ in $E$ and two roots $\pm b$ outside of $E$.]

Now for the key point: given a general finite field extension $L : K$, it is not true that $|{\rm Gal}(L : K)| = [L:K]$! This equality holds if and only if the extension $L : K$ is normal and separable, but in our example, $E : \mathbb Q$ is not normal.

[The correct general statement is that $|Gal(L:K)| = [L : \Phi({\rm Gal}(L:K))]$, where $\Phi({\rm Gal}(L:K))]$ is the subfield of $L$ consisting of those elements that are fixed by all automorphisms in ${\rm Gal}(L:K))]$. Only in the case where $L : K$ is normal and separable do we have $\Phi({\rm Gal}(L:K)) = K$.]

Anyway, you're right that $|{\rm Gal}(E : \mathbb Q)| = 2$. Since $E = \mathbb Q (a)$, every automorphism in ${\rm Gal}(E : \mathbb Q)$ is uniquely determined by where it sends $a$. As you pointed out, the image of $a$ under any automorphism in ${\rm Gal}(E : \mathbb Q)$ must be either $\pm a$. This is because automorphisms in ${\rm Gal}(E : \mathbb Q)$ must map roots of $x^4 - 2x^2 - 1$ to roots of $x^4 - 2x^2 - 1$. Furthermore, there really does exist an automorphism sending $a \mapsto -a$. (This is more obvious if you think of $\mathbb Q(a)$ as $\mathbb Q[x]/(x^4 - 2x^2 - 1)$, with $x$ representing $a$.)

Finally, let's check that the formula $|Gal(E:\mathbb Q)| = [E : \Phi({\rm Gal}(E:\mathbb Q))]$ holds here. The subfield $\Phi({\rm Gal}(E:\mathbb Q))$, which by definition consists of all elements of $E$ fixed under $a \mapsto \pm a$, is the subfield $\mathbb Q(a^2) = \mathbb Q(\sqrt 2)$. And yes, $[E : \mathbb Q(\sqrt 2)] = 2 = |{\rm Gal}(E : \mathbb Q)|$, as expected.

• This is interesting, because we're calling the group $Gal(E/ \mathbb{Q})$, but it isn't actually a Galois group is it? Because if it was, then $E$ would be a splitting field, but based off of what you've described, $E$ is not a splitting field of the minimal polynomial over $\mathbb{Q}$. This makes me wonder, is my minimal polynomial correct? Thanks so much for your input. This discussion is much more interesting than I'd expected. – obewanjacobi Jul 27 '17 at 19:31
• @obewanjacobi That's interesting! It appears that different people use different notation and different terminology, hence the confusion! In Howie's "Fields and Galois theory", $Gal(L:K)$ denotes the group of automorphisms of $L$ fixing $K$, regardless of whether $L:K$ is normal or not. This is also mentioned in a short remark in Dummit & Foote. But on Wikipedia, $Gal(L:K)$ refers to the automorphism group of the normal extension of $L$ over $K$ (which in this case would be the splitting field); the automorphism group of $L$ fixing $K$ is instead denoted by $Aut(L:K)$. – Kenny Wong Jul 27 '17 at 21:37
• @obewanjacobi But yes, your minimal polynomial is correct, and it is up to you whether you interpret the question as asking for $Aut(\mathbb Q(a) : \mathbb Q)$ (this has order two) or as asking for $Aut(\mathbb Q(a,b) : \mathbb Q)$ (this has order eight). – Kenny Wong Jul 27 '17 at 21:39
• Understood. Thanks so much for your input on this problem. This came from an old qualifying exam from my University, so I will check with one of the professors there which way they would prefer it be interpreted. – obewanjacobi Jul 27 '17 at 21:53

In this case the splitting field is generated by $\sqrt{1+\sqrt2}$ and $\sqrt{1-\sqrt2}$. These generate a degree $4$ extension of $\Bbb Q(\sqrt2)$ (why?). There will be Galois automorphisms taking $\sqrt{1+\sqrt2}$ to $\pm\sqrt{1-\sqrt2}$.

• I guess the arithmetic is wrong, $\sqrt{1+\sqrt2}\cdot\sqrt{1-\sqrt2}=\sqrt{1-2}=i$. – Taisuke Yasuda Jul 26 '17 at 18:11
• @TaisukeYasuda Thanks! – Kenny Wong Jul 26 '17 at 18:11