# How to formally prove that the degree of $\Bbb Q（\sqrt[5]{6},\sqrt[3]{7}）$ is $15$？

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！

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)

• 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？ – non-abelian group of order 9 Apr 13 '17 at 11:10
• 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 – Silvia Apr 13 '17 at 11:15

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