# 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.

• More generally, any quadratic extension is normal. – carmichael561 Feb 20 at 23:59
• @carmichael561: Hear, hear! :) – Robert Lewis Feb 21 at 0:07

$$x^2-5$$ 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.

• Cool, that's what I thought! Thanks so much. Glad that my understanding wasn't totally off. Also, I notice that my post just got reformatted to look a lot nicer. Is this automated, or did some kind of moderator go and do that by hand? – Chris N-L Feb 20 at 23:19
• People can propose edits, and with enough reputation can just make them happen. (You should really learn MathJax if you intend to post more questions or answers on this site. Try here. – rogerl Feb 20 at 23:21
• Awesome, will do. Thank you. – Chris N-L Feb 20 at 23:31

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.)