2
$\begingroup$

I tried to consider the tower of extension $\Bbb Q\subset \Bbb Q(\sqrt[3]{7})\subset\Bbb Q(\sqrt[5]{6},\sqrt[3]{7})$.

The minimal polynomial of $\Bbb Q(\sqrt[3]{7})$ over $\Bbb Q$ is $x^3-7$ by Eisenstein. But although it is easy to see that it has no root in $\Bbb Q(\sqrt[3]{7})$, how can I formally conclude that $x^5-6$ is irreducible over $\Bbb Q(\sqrt[3]{7})$ and thus we can see the basis of $\Bbb Q(\sqrt[3]{7}, \sqrt[5]{6})$?

Thanks!

$\endgroup$
3
$\begingroup$

Hint: Consider both extensions $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{7})\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ and $\mathbb{Q}\subset\mathbb{Q}(\sqrt[5]{6})\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$. Which conditions (multiple of...) must the degree of $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ satisfy (using transitivity of degree)?

(Edit: in other words, draw the two towers next to each other)

$\endgroup$
  • $\begingroup$ I can see the fact that as 3 and 5 are coprime, the degree of the extension on the top of the tower is at least 15 since it must be divisible by both 3 and 5. But may I ask what to do next to conclude that it has degree 15? $\endgroup$ – non-abelian group of order 9 Apr 13 '17 at 11:10
  • $\begingroup$ Good! Well, as you pointed out, $x^{5}-6$ has $\sqrt[5]{6}$ as a root, but we cannot (in principle) tell if this is irreducible. However, we can deduce from it that the minimal polinomial must at least divide $x^{5}-6$, right? In that case, the degree of $\mathbb{Q}(\sqrt[3]{7})\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ must be at most 5, true? But then the degree of $\mathbb{Q}\subset\mathbb{Q}(\sqrt[3]{7},\sqrt[5]{6})$ must (by transitivity) be at most 15, right? But you already got that it had to be at least 15... so it's 15 $\endgroup$ – Silvia Apr 13 '17 at 11:15
2
$\begingroup$

Since $\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)$ is a finitely generated algebraic extension of $\mathbb{Q}$, it is a finite extension of $\mathbb{Q}$.

Hence $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)]=[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[5]6)][\mathbb{Q}(\sqrt[5]6):\mathbb{Q}]$.

And also, $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7)]=[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[3]7)][\mathbb{Q}(\sqrt[3]7):\mathbb{Q}]$.

You are correct in your reasoning for $[\mathbb{Q}(\sqrt[3]7):\mathbb{Q}]=3$ and $[\mathbb{Q}(\sqrt[5]6):\mathbb{Q}]=5$.

Now consider $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}(\sqrt[5]6)]$.

You will find that the minimum polynomial of $\sqrt[3]7$ over $\mathbb{Q}$, $m(x)$ say, divides $x^3-7$ in $\mathbb{Q}(\sqrt[5]6)$.

Let $deg(m(x))=k$. Then $1\leq k \leq 3$.

Then from the second line in this answer, we then have:

$[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}]=5k$.

Now from the third line we find, $3|5k$.

But by Euclid's Lemma, $gcd(5,3)=1$. So $3|k$.

But $1\leq k \leq 3$, hence $k=3$.

Thus $[\mathbb{Q}(\sqrt[5]6,\sqrt[3]7):\mathbb{Q}]=3.5=15$.

$\endgroup$

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.