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In an assignment I got I have been asked to try to determine when $E/F$ is a Galois Extension, and determine the Galois group of such an extension.

$\textbf{Context:}$ $F$ is any field of characteristic zero and $E = F(\sqrt{c},\sqrt{a + b\sqrt{c}})$, where $a,b,c \in F$, $c$ is not a square in $F$ and $a + b\sqrt{c}$ is not a square in $F(\sqrt{c})$. Now assume that $b \neq 0$; I have determined that this is a Galois extension iff

$$\sqrt{ a -b\sqrt{c}}$$ is in $E$. This holds iff $a - b\sqrt{c}$ is the square of an element in $F(\sqrt{c})$, or $(a + b\sqrt{c})(a - b\sqrt{c} ) = a^2 - b^2c$ is the square of an element in $F(\sqrt{c})$. Now the first case cannot hold because it contradicts our assumption that $a + b\sqrt{c}$ is not a square in $F(\sqrt{c})$. So the second case holds, that is when $a^2 -b^2c = (h + g\sqrt{c})^2$ for some $h,g \in F$.

Expanding this out and comparing coefficients, it must be the case that either $h = 0$ or (exclusively) $g = 0$. The first case gives that

$$a^2 - b^2c = g^2c$$ for some $g \in F$, or (exclusively)

$$a^2 - b^2c = h^2$$ for some $h \in F$.

$\textbf{Where I'm stuck:}$ Now the problem for me comes in determining the Galois group of $E/F$. Firstly $E$ can be written as $F(\sqrt{a + b\sqrt{c}}, \sqrt{a - b\sqrt{c}})$ or even just $F(\sqrt{a + b\sqrt{c}})$.

If we take $E = F(\sqrt{a + b\sqrt{c}}, \sqrt{a - b\sqrt{c}})$ then noticing that $E$ is the splitting field of $(x^2 - a)^2 - b^2c$ over $F$, my guess is that $\sigma \in \operatorname{Gal}(E/F)$ can only take $\sqrt{a + b\sqrt{c}}$ to $- \sqrt{a + b\sqrt{c}}$, it can't take $\sqrt{a + b\sqrt{c}}$ to say $\sqrt{a - b\sqrt{c}}$. I am guessing this based on looking at the polynomial $(x^2 - a)^2 - b^2c$ which can be written as

$$(x^2 - (a + b\sqrt{c}))(x^2 - (a - b\sqrt{c})).$$

However it seems to me that I am not taking advantage of the conditions found for $a^2 - b^2c$ above. In addition how can I determine whether given some $\sigma$ a permutation of the roots, that it is actually a valid member of the Galois group $\operatorname{Gal}(E/F)$?


$\textbf{Edit:}$ I think the second case where $a^2 -b^2c = h^2$ cannot hold because this would contradict the fact that the degree of $E = F(\sqrt{a +b\sqrt{c}},\sqrt{a - b\sqrt{c}})$ over $F$ is 4.

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Maybe it would help if you worked your way through an example, such as ${\bf Q}(\sqrt2,\sqrt{2+\sqrt2})$. – Gerry Myerson May 5 '12 at 6:41
@GerryMyerson I have tried your example, and that corresponds to the first case where $a^2 - b^2c = g^2c$ for some $g \in F$. In your example we have $g = 1 \in \Bbb{Q}$. However this is not helping me at all because I am still coming back to my confusions above. – user38268 May 5 '12 at 6:48
@GerryMyerson I think I am going somewhere: I think the case where $a^2 - b^2c = h^2$ cannot hold, while in the first case when $a^2 - b^2c = g^2c$ we always get the Klein 4 group...... – user38268 May 5 '12 at 6:54
@JyrkiLahtonen I don't think I mentioned before that I need $z$ to be in $F(\sqrt{c})$. I mentioned that I needed it to be in $E$. – user38268 May 5 '12 at 7:33
@JyrkiLahtonen Actually I think I have resolved one case, the case when $a^2 - b^2c = h^2$. Suppose in this case we have an element of order $4$, $\sigma$ in the Galois group. Then wlog $\sigma(\sqrt{a + b\sqrt{c}}) = \sqrt{a - b\sqrt{c}}$. Now since this has order 4 we must have that $\sigma(\sqrt{a - b\sqrt{c}}) = - \sqrt{a + b\sqrt{c}}$. But then if we multiply these together we get that $\sigma(\sqrt{a^2 - b^2c}) = - \sqrt{a^2 - b^2c}$. But $\sqrt{a^2 - b^2c}$ is $\pm h$ that is an element of $F$ that must be fixed by $\sigma$. This is a contradiction. So in this case (continued) – user38268 May 5 '12 at 8:06
up vote 3 down vote accepted

Benjamin has done nearly all the work. I try to clean up some points.

From the assumptions it follows that $z=\sqrt{a+b\sqrt c}$ generates a quartic Galois extension $E=F(z)$. Its conjugates are $z_1=z$, $z_2=\sqrt{a-b\sqrt c}$, $z_3=-z_1$ and $z_4=-z_2$. Let $\sigma_j$ be the unique $F$-automorphism such that $\sigma_j(z_1)=z_j, j=1,2,3,4$ (such automorphisms exist, because the Galois group permutes the conjugates transitively, and they are unique, because $E=F(z_1)$). We can also identify the automorphisms $\sigma_j\in G=Gal(E/F)$ as elements of $S_4=Sym(z_1,z_2,z_3,z_4)$ in the usual way. Observe that only the identity element $\sigma_1$ can have fixed points among the roots. This is because $E=F(z_j)$ for all $j$, so an automorphism fixing any root $z_j$ is necesssarily the identity mapping.

The automorphism $\sigma_3$ maps $\sqrt c=(z_1^2-a)/b$ to $(z_3^2-a)/b=\sqrt c$, so its fixed field contains $F(\sqrt c)$, and consequently $\sigma_3$ has order two. We can conclude that $\sigma_3=(13)(24)$.

The answer hinges on the question: What is the order of $\sigma_2$? Obviously it is either two or four.

Assume first that $\sigma_2$ has order two. In that case $\sigma_2(z_2)=z_1$, and consequently $\sigma_2=(12)(34)$. Therefore $\sigma_4=\sigma_2\circ\sigma_3=(14)(23)$, and $G$ is the Klein 4-group. Write $w=z_1+z_2$. Here $\sigma_2(w)=w$, and $$ \sigma_4(w)=\sigma_4(z_1)+\sigma_4(z_2)=z_3+z_4=-w=\sigma_3(w). $$ Thus the minimal polynomial of $w$ over $F$ is $$(x-w)(x-\sigma_4(w))=(x-w)(x+w)=x^2-w^2=x^2-(2a+2\sqrt{a^2-b^2c}).$$ The coefficients of this polynomial must be in $F$, so we can conclude that we are in Benjamin's case $g=0, h=\sqrt{a^2-b^2c}\in F$. In this case the fixed field of $\sigma_2$ is thus $F(\sqrt{2(a+h)}$. As an aside we observe that a similar calculation shows that in this case the fixed field of $\sigma_4$ is $F(\sqrt{2(a-h)})=F(z_1+z_4)$.

Assume then that $\sigma_2$ has order four, or equivalently that $G$ is cyclic. Then $\sigma_3$ is the only element of order two in $G$, so $\sigma_2^2=\sigma_3$, and thus $\sigma_2=(1234)$. Let's again write $w=z_1+z_2$. Again $\sigma_3(w)=-w$. However, this time $w$ is not in the fixed field of $\sigma_2$. Therefore we can only conclude that the coefficients of $$ (x-w)(x+w)=x^2-(2a+2\sqrt{a^2-b^2c}) $$ are in the fixed field of $\sigma_3$. But earlier we identified the fixed field of $\sigma_3$ to be $F(\sqrt c)$. Thus we are in Benjamin's case $a^2-b^2c=g^2c, g\in F$.

For the sake of completeness let us show that both cases occur by way of examples. In both cases I use $F=\mathbf{Q}$. The choices $c=2, a=2, b=1$, lead to the cyclic case (as was evidently observed by Gerry Myerson). Here $E=F(\sqrt{2+\sqrt2})$ is the real subfield of the cyclotomic field $F(\zeta_{16})$. The Klein quartic is obtained with the choices $c=3, a=2, b=1$. In that case $E=F(\sqrt{2+\sqrt3})$ is the real subfield of $F(\zeta_{24})$.

The Galois groups can be identified as quotient groups of the Galois groups of cyclotomic fields. In the first case we get $$ G=\mathbf{Z}_{16}^*/\langle -1\rangle=\{1,3,5,7,9=-7,11=-5,13=-3,15=-1\}/\langle -1\rangle $$ that is generated by the coset $3$. In the latter case we have $$ G=\mathbf{Z}_{24}^*/\langle -1\rangle=\{1,5,7,11,13=-11,17=-7,19=-5,23=-1\}/\langle -1\rangle $$ that is isomorphic to the Klein group as $5^2\equiv7^2\equiv11^2\equiv1\pmod{24}$.

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A way of identifying the example cases as the real subfields of the claimed cyclotomic fields is to calculate $$ \cos\frac\pi{8}=\sqrt{\frac{1+\cos\pi/4}2}=\frac12\sqrt{2+\sqrt2} $$ and $$ \cos\frac\pi{12}=\sqrt{\frac{1+\cos\pi/6}2}=\frac12\sqrt{2+\sqrt3} $$ – Jyrki Lahtonen May 5 '12 at 10:37
I am very thankful to you for spending your time to help me. – user38268 May 5 '12 at 10:51

Since you have already established that if $a^2 - b^2c = h^2$ for rational $h$, that $\mathrm{Gal}(E/F) \cong V$, I will only address the "other case" where $a^2 - b^2c = g^2c$ for some rational $g$.

It suffices to show that $\sigma:\sqrt{a+b\sqrt{c}} \mapsto \sqrt{a-b\sqrt{c}}$ is of order $4$. Let $\alpha = \sqrt{a+b\sqrt{c}}$ and $\beta = \sqrt{a-b\sqrt{c}}$. Then $\sigma$ is of order $4$ iff $\sigma(\beta) \neq \alpha$.

Note that $\alpha\beta = (g/b)(\alpha^2 - a)$, so $\beta = (g/b)(\alpha - (a/\alpha))$. Hence:

$\sigma(\beta) = (g/b)(\beta - (a/\beta)) = (g/b)((\beta^2 - a)/\beta) = (g/b)(-b\sqrt{c}/\beta) = (-g\sqrt{c}/\beta)$ $ = (-\alpha\beta/\beta) = -\alpha$,

so $\sigma$ is of order 4, so in this case $\mathrm{Gal}(E/F) \cong C_4$.

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Thanks so much for your post. However I found a much simpler way of showing that this must have order 4: Observe that since $\sigma(\alpha) = \beta$, then $\sigma(\alpha^2) = \beta^2$ which means that $a + b\sigma(\sqrt{c}) = a - b\sqrt{c}$ and hence that $\sigma(\sqrt{c}) = -\sqrt{c}$. From there we get immediately that $\sigma(\beta) = \sigma(g\sqrt{c}/\alpha) = -g\sqrt{c}/\beta = -\sqrt{a^2 - b^2c}/\beta = -\alpha$. – user38268 May 5 '12 at 13:45
we get from $a + b\sigma(\sqrt{c}) = a-b\sqrt{c}$ to $\sigma(\sqrt{c}) = -\sqrt{c}$ because we can cancel $b$ since $ b\neq 0$. – user38268 May 5 '12 at 13:54
quite so. the thing to realize is that $\sigma$ is the conjugation map on $F(\sqrt{c})$, as someone pointed out above. – David Wheeler May 5 '12 at 13:55
Man you freaked me out with that manipulation of yours! – user38268 May 5 '12 at 13:56

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