Is $[\Bbb Q(5^{1/2}, 5^{1/7}): \Bbb Q(5^{1/7})]$ a normal extension? I've working on a problem set with a bunch of these, and I get the idea generally. An extension is normal if all of the roots for the min pol for the element we are extending by are in the field. e.g., $\Bbb Q(2^{1/3})$ is not, but it is if we add $\omega$, a root of unity. 
I don't really understand how to consider this scenario though. To start, $[\Bbb Q(5^{1/7}):\Bbb Q]$ is not normal. But, if our 'base' field isn't $\Bbb Q$, but $Q(5^{1/7})$, and we append $5^{1/2}$, the roots of the min pol for $5^{1/2}$ are in the extension. Herein lies my confusion. My hypothesis is that it is in fact a normal extension. 
 A: $x^2-5$ is certainly a polynomial over $\mathbb{Q}(5^{1/7})$ of which $5^{1/2}$ is a root, and clearly the other root of $x^2-5$ is also in $\mathbb{Q}(5^{1/2},5^{1/7})$. The minimal polynomial of $5^{1/2}$ over $\mathbb{Q}(5^{1/7})$ must therefore divide $x^2-5$, so it is either $x^2-5$ or a linear polynomial. So in either case the extension is normal.
A: I think a fact which helps out here is that quadratic field extensions are always normal; that is, if $F$ is any field, and $E$ is an extension field such that $[E:F] = 2$, then $E/F$ is normal over $F$; for if
$\alpha \in E \setminus F, \tag 1$
then there exists a monic quadratic polynomial 
$q(x) \in F[x], \tag 2$
such that
$q(\alpha) = 0; \tag 3$
writing
$q(x) = x^2 + ax + b, \; a, b \in F,\tag 4$
and dividing $q(x)$ by $x - \alpha$ we find
$q(x) = (x - \alpha)(x + (a + \alpha)), \tag 5$
which may easily be checked:
$(x - \alpha)(x + (a + \alpha)) = x^2 - \alpha x + (a + \alpha)x - \alpha(a + \alpha) = x^2 + ax - \alpha^2 - a \alpha, \tag 6$
and it follows from (3) and (4) that
$\alpha^2 + a \alpha + b = 0 \Longrightarrow b = -\alpha^2 - a\alpha; \tag7$
thus (6) becomes
$(x - \alpha)(x + (a + \alpha)) = x^2 + ax + b = q(x); \tag 8$
it follows then that, given (3), the roots of $q(x)$ are
$\alpha, -a - \alpha \in E; \tag 9$
since $q(x)$ splits in $E[x]$, $E/F$ is a normal extension.
Applying this little result to the case at hand, we see that since
$x^2 - 5 \in \Bbb Q(5^{1/7})[x] \tag{10}$
we have
$[\Bbb Q(5^{1/7}, 5^{1/2}):\Bbb Q(5^{1/7}] \le 2, \tag{11}$
so the extension $\Bbb Q(5^{1/7}, 5^{1/2}) / \Bbb Q(5^{1/7})$ is normal (here we take into account that
$[E:F] = 1 \Longrightarrow E/F \; \text{is normal}, \tag{12}$
always).  In fact
$[\Bbb Q(5^{1/7}, 5^{1/2})/\Bbb Q(5^{1/7})] = 2, \tag{13}$
since $x^7 - 5$ is irreducible over $\Bbb Q$ by Eisenstein with $p =  5$; thus
$[\Bbb Q(5^{1/7}): \Bbb Q] = 7, \tag{14}$
so if
$5^{1/2} \in \Bbb Q(5^{1/7}), \tag{15}$
then
$\Bbb Q(5^{1/2}) \subset \Bbb Q(5^{1/7}), \tag{16}$
and then
$[\Bbb Q(5^{1/7}):\Bbb Q] = [\Bbb Q(5^{1/7}):\Bbb Q(5^{1/2})] [\Bbb Q(5^{1/2}): \Bbb Q] \Longrightarrow 2 \mid 7, \tag{17}$
impossible!  (And here of course we use the fact that $[\Bbb Q(5^{1/2}): \Bbb Q] = 2$ since $x^2 - 5$ is irreducible over $\Bbb Q$, also by Eisenstein.)
