Compute $|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}|$

My attempt so far :

$|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}|$ $=|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})|$.$|\mathbb Q(\sqrt[5] {3}):\mathbb{Q}|$


$|\mathbb Q(\sqrt[5] {3}):\mathbb{Q}|=5$

Now i need to calculate:

$|\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})|$

Which i think is $2$ but i am unsure to calculate . Is this the best method and if so how do i finish the problem?

Also how do i find a bases of $\mathbb Q(\sqrt {2},\sqrt[5] {3})$ over $\mathbb Q $ ?


$[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]$ is a multiple of both $2$ and $5$ because $$ [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}] = [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt {2})] \,[\mathbb Q(\sqrt {2}):\mathbb{Q}] = 2[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt {2})] $$ $$ [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}] = [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})] \, [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}] =5[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})] $$

On the other hand, $$ [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})] = [\mathbb Q(\sqrt[5] {3})(\sqrt {2}):\mathbb Q(\sqrt[5] {3})]\le 2 $$ because $\sqrt {2}$ is a root of quadratic polynomial with coefficients in $\mathbb Q \subseteq \mathbb Q(\sqrt[5] {3})$, and so $$ [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}] = [\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})] \, [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}] \le 2 [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}] = 2 \cdot 5 = 10 $$ Therefore, $[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb{Q}]=10$.

  • 1
    $\begingroup$ $$[\mathbb Q(\sqrt {2},\sqrt[5] {3}):\mathbb Q(\sqrt[5] {3})] \, [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}] \le 2 [\mathbb Q(\sqrt[5] {3}):\mathbb{Q}]$$ I am having trouble understanding why this is so $\endgroup$ – Tim Jones Apr 1 '18 at 17:25
  • 1
    $\begingroup$ @TimJones, see my edited answer. $\endgroup$ – lhf Apr 1 '18 at 21:19

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