1
$\begingroup$

I've just started field theory from Gallian. It may happen that the reasoning I'm providing for this problem seems weird. But this is all by what I tried to learn from some problems on maths stack exchange.

Suppose $\mathbb F \subset \mathbb C$ is the splitting field of $x^7-2$ over $\mathbb Q$ and $z=e^{\frac {2\pi}{7}}$ a primitive seventh root of unity. Let [$\mathbb F: \mathbb Q(z)]=a$ and $[\mathbb F: \mathbb Q(2^{(1/7)})]=b$. Then

(A) $a=b=7$

(B) $a=b=6$

(C) $a>b$

(D) $a<b$

Since degree of field extension is the degree of irreducible polynomial whose root is $e^{\frac {2\pi}{7}}$ and the coefficient is in $\mathbb Q(2\pi/7)$. Here the polynomial is $p(z)=z^7-1$. So $a=7$,similarly $b=7$ but this wrong. I don't know the correct answer.

If possible please explain me in detail how to find the degree of field extension, splitting fields for some polynomial. What things we have to take care of while finding the degree of field extension, splitting fields for some polynomial?

$\endgroup$
  • 4
    $\begingroup$ Note that $x^7-1$ is not irreducible: $x^7-1 = (x-1)(x^6+\ldots+x+1)$, and $1 \in \mathbb Q$. $\endgroup$ – flawr Jun 5 '17 at 10:17
  • $\begingroup$ ... but $x^7 - 2$ is irreducible over $\mathbb{Q}$, e.g. by Eisenstein's Criterion. $\endgroup$ – hardmath Jun 8 '17 at 16:55
2
$\begingroup$

First, you have a typo, $\large z = e^{\frac{2\pi}{7}i}$. Anyway, note that $\mathbb{F} = \mathbb{Q}(z)(2^{\frac{1}{7}}).$ So $[\mathbb{F} : \mathbb{Q}(z)]$ is the degree of the minimal polynomial of $2^{\frac{1}{7}}$ over the field $\mathbb{Q}(z)$. The minimal polynomial is still $x^7-2$, so $[\mathbb{F} : \mathbb{Q}(z)] = 7.$

Now, alternatively $\mathbb{F} = \mathbb{Q}(2^{\frac{1}{7}})(z)$. So $[\mathbb{F} : \mathbb{Q}(2^{\frac{1}{7}})]$ is the degree of the minimal polynomial of $z$ over $\mathbb{Q}(2^{\frac{1}{7}})$. The minimal polynomial of $z$ is $\textbf{not}$ $x^7-1$, this is not irreducible. The minimal polynomial of $z$ is $1 + x + x^2 + x^3 + x^4 + x^5 + x^6$ (for a discusion of minimal polynomials of primitive roots of unity, check out the Wikipedia page on Cyclotomic Polynomials.) Anyway, we see that $[\mathbb{F} : \mathbb{Q}(2^{\frac{1}{7}})] = 6$.

Side note: Though this is not asked in the question, we can see that $$[\mathbb{F} : \mathbb{Q}] = [\mathbb{F}: \mathbb{Q(z)}][\mathbb{Q}(z):\mathbb{Q}]= 7*6 = 42.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.