Is the extension $\mathbb{Q}(\sqrt{7+\sqrt{7}})/\mathbb{Q}$ a Galois extension?

Question

I'm preparing for an algebra exam, and in previous exams I found the following exercise:

Let $\alpha = \sqrt{7+\sqrt{7}}$. Show that $\mathbb{Q}(\alpha)$ is a Galois extension of $\mathbb{Q}$ of degree $4$ with cyclic Galois group.

Now, my problem is that this extension seems not to be a normal extension. Here is my reasoning: It is easy to see that the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $$ f(X) = X^4-14X^2+42 $$ which is irreducible by Eisenstein criterion. The roots of $f(X)$ are $\pm\alpha$ and $\pm\sqrt{7-\sqrt{7}}$, so the splitting field of $f(X)$ is $$ K = \mathbb{Q}\left(\alpha,\sqrt{7-\sqrt{7}}\right). $$ Note that $\sqrt{7}\in \mathbb{Q}(\alpha)\subseteq K$ and because $$ \alpha\sqrt{7-\sqrt{7}} = \sqrt{42} = \sqrt{7}\sqrt{6}, $$ we have that $\sqrt{6}\in K$. Also from this last equation if follows that $\sqrt{7-\sqrt{7}}\in \mathbb{Q}[\alpha,\sqrt{6}]$, thus $K=\mathbb{Q}(\alpha,\sqrt{6})$. Now we have a tower of field extensions $$ \mathbb{Q}\subseteq F=\mathbb{Q}[\sqrt{6}] \subseteq K. $$ The polynomial $f(X)$ is irreducible over $F$: Indeed, it has no roots in $F$ because this would imply that $\sqrt{7}\in F$ which is easy to see to be impossible. Thus if $f(X)$ where reducible over $F$ it must be the product of two polynomials of degree $2$ over $F$, that is $$ f(X) = (X^2+aX+b)(X^2+cX+d), \quad a,b,c,d\in F. $$ Expanding and comparing coefficients, we obtain $$ a+c = 0, \quad b+d+ac=-14, \quad ad+bc=0, \quad bd=42, $$ from where $$ c=-a, \quad b+d-a^2=-14, \quad a(b-d)=0, \quad bd=42. $$ The third equation in the above line tells us that $a=0$ or $b=d$. If $a=0$ we obtain that $$ b+d=-14 \quad \text{and} \quad bd=42, $$ and solving for $b$ we obtain $b=7\pm\sqrt{7}\not\in F$, a contradiction. If $b=d$ then $b^2=42$, giving $b=\pm \sqrt{6}\sqrt{7}\not\in F$. Thus $f(X)$ must be irreducible over $\mathbb{Q}[\sqrt{6}]$.

This implies that $\mathbb{Q}(\alpha)$ is not the splitting field of $f(X)$ and hence $\mathbb{Q}(\alpha)/\mathbb{Q}$ is not a normal extension.

From here I know how to proceed to compute the Galois group, but I want to be sure that I made no mistakes. So my question is: Is this reasoning correct? Did I made a mistake or was it a mistake of the professor?

Answer 1

Yeah! that's perfect. For your conviction this is what magma says:

P< x >:=PolynomialAlgebra(Rationals());

f:=x^4-14*x^2+42;

G:=GaloisGroup(f);

print G;

Permutation group G acting on a set of cardinality 4 Order = 8 = 2^3 (1, 2)(3, 4) (1, 4)

If $\mathbb{Q}(\alpha)|\mathbb{Q}$ were Galois then $[\mathbb{Q}(\alpha):\mathbb{Q}]$ would be 8, but we know it's 4

Answer 2

If the Galois groups is isomorphic to $\mathbb Z/4\mathbb Z$, then it can only have 1 proper intermediate extension, but from your work, we know both $\sqrt{6}$ and $\sqrt{7}$ are in the normal closure of $\mathbb Q[\alpha]$.

To show the polynomial is irreducible or equivalently $[\mathbb Q[\alpha]:\mathbb Q]=4$, we have the following more conceptual way. As clearly $\sqrt{7}\in\mathbb Q[\alpha^2]\subset\mathbb Q[\alpha]$, and $[\mathbb Q(\sqrt 7):\mathbb Q]$ divides $[\mathbb Q(\alpha):\mathbb Q]\le 4$, hence $[\mathbb Q[\alpha]:\mathbb Q]$ can only be $2$ or $4$. If it is $2$, then $\mathbb Q[\alpha]=\mathbb Q[\sqrt 7]$.

Now let $\sigma$ be the nontrivial element in $\text{Gal}(\mathbb Q(\sqrt 7)/\mathbb Q)$, we have $$[\sigma(\sqrt{7+\sqrt 7})]^2=\sigma(7)+\sigma(\sqrt 7)=7-\sqrt 7$$ $$\sigma(\sqrt{7+\sqrt 7})=\pm \sqrt{7-\sqrt 7}$$ And $\mathbb Q(\sqrt 7)$ will contain $\sqrt{7-\sqrt 7}$ no matter what the sign is. Further it also contains $(\sqrt{7+\sqrt 7})(\sqrt{7-\sqrt 7})=\sqrt{42}$ and $\frac{\sqrt{42}}{\sqrt 7}=\sqrt 6$ and it's not hard to show $\sqrt 6\not\in\mathbb Q(\sqrt 7)$.