$K=\Bbb Q(\sqrt3,\sqrt[3]{2}) $, Compute $[K:\Bbb Q]$.

Question

I've tried looking at similar examples for but something has been close enough. How do I compute $[\mathbb Q(\sqrt 3,\sqrt[3]2):\mathbb Q]$.

I think you say that it's equal to $[\mathbb Q(\sqrt3 + \sqrt[3]2):\mathbb Q]$, and I think the answer is $6$ but I don't knows how to show it.

Also do I find the minimal polynomial for this over $\mathbb Q$?

Answer 1

By Tower law, we have $[\Bbb Q (\sqrt3,\sqrt[3]{2}):\Bbb Q]=[\Bbb Q (\sqrt[3]{2})(\sqrt3):\Bbb Q(\sqrt[3]{2})]\cdot [\Bbb Q (\sqrt[3]{2}):\Bbb Q] .$ Clearly, $[\Bbb Q (\sqrt[3]{2}):\Bbb Q]=3$. Also note that the minimal polynomial of $\sqrt3$ over $\Bbb Q(\sqrt[3]{2})$,say $p (x) $, divides $x^2-3$. So degree of $p (x) \leq 2$.

Suppose degree is $1$. Then $\sqrt3\in\Bbb Q (\sqrt[3]{2})$. So $\Bbb Q(\sqrt3)\subseteq\Bbb Q (\sqrt[3]{2})$. Now again by tower law, $$3=[\Bbb Q (\sqrt[3]{2}):\Bbb Q]\\=[\Bbb Q (\sqrt[3]{2}):\Bbb Q(\sqrt{3})]\cdot [\Bbb Q (\sqrt{3}):\Bbb Q]\\=[\Bbb Q (\sqrt[3]{2}):\Bbb Q(\sqrt{3})]\cdot2,$$ that is, $3=[\Bbb Q (\sqrt[3]{2}):\Bbb Q(\sqrt{3})]\cdot2,$ a contradiction.

Answer 2

You are right, $[\Bbb{Q}(\sqrt3, \sqrt[3]2 ): \Bbb{Q}]=6$ .

You need to compute $[\Bbb{Q}( \sqrt3 ): \Bbb{Q}]$ and $[\Bbb{Q}(\sqrt3, \sqrt[3]2 ): \Bbb{Q}(\sqrt3)]$. Then use tower law.

The minimal polynomials are $\mu_{\sqrt3, \Bbb{Q}}=X^2-3$ and $\mu_{\sqrt[3]2, \Bbb{Q}(\sqrt3)}=X^3-2$. You need to show that those are irreducible over the fields given.

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A general method to find the minimal polynomial of $q$ over a given field is to set $X=q$ and transform the equation by squaring/cubing/etc. until you get an equation of the form $p(X)=0$, where $p(X)$ is a polynomial over our field and then checking whether it is irreducible. This can get very messy though.

Please note that the resulting polynomial MAY NOT BE IRREDUCIBLE, but the minimal polynomial will be one of its factors.

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Example with $\sqrt3+ \sqrt[3]2$:

$X=\sqrt3+ \sqrt[3]2 \iff X-\sqrt3= \sqrt[3]2 \iff (X-\sqrt3)^3=2 \iff$

$X^3- 3 \sqrt3 X^2 + 9 X- 3 \sqrt3=2 \iff X^3 +9X -2 = 3 \sqrt3 X^2+3 \sqrt3 \iff$

$(X^3 +9X -2)^2 = (3 \sqrt3 X^2+3 \sqrt3)^2 \iff$

$X^6 + 18 X^4 - 4 X^3 + 81 X^2 - 36 X + 4 = 27 X^4 + 54 X^2 + 27 \iff $

$X^6 - 9 X^4 - 4 X^3 + 27 X^2 - 36 X - 23=0$

This polynomial is normed and you can check that it is irreducible by using transformations, checking irreducibility modulo primes, or by factoring it over $\Bbb{R, C}$ and seeing that no factor product of the polynomial is in $\Bbb{Q}$.

Answer 3

We have $[\Bbb{Q}( \sqrt3 ): \Bbb{Q}]=2$ and $[\Bbb Q (\sqrt[3]{2}):\Bbb Q]=3$.

Therefore, $[\Bbb Q (\sqrt3,\sqrt[3]{2}):\Bbb Q] \ge 6$ because it is a multiple of both $2$ and $3$ by the tower law.

On the other hand, $$ [\Bbb Q (\sqrt3,\sqrt[3]{2}):\Bbb Q] =[\Bbb Q (\sqrt[3]{2})(\sqrt3):\Bbb Q(\sqrt[3]{2})]\cdot [\Bbb Q (\sqrt[3]{2}):\Bbb Q] \le [\Bbb Q(\sqrt3):\Bbb Q]\cdot [\Bbb Q (\sqrt[3]{2}):\Bbb Q] =6 $$