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 $\mathbb Q$?


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


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.

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.

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 $\Bbb{Q}$.

  • $\begingroup$ Sorry I wasn't clear on the minimal polynomial part. I've got to find the minimal polynomial of ((√3)+2^(1/3)) over Q. $\endgroup$ – Freddie Mar 21 at 14:24
  • $\begingroup$ I will edit my answer with this example. $\endgroup$ – B.Swan Mar 21 at 16:51
  • $\begingroup$ And as an extension, if I wanted to find a basis of K over Q, how do you do that in this example? $\endgroup$ – Freddie Mar 22 at 12:34
  • $\begingroup$ @Freddie What do the elements of K look like? $\endgroup$ – B.Swan Mar 22 at 13:32
  • 1
    $\begingroup$ @ThomasShelby I said "not obviously irreducible" which is different from "obviously not irreducible". The first means it maybe is irreducible, but it is not clear at first sight. The second that it is clear that its reducible. The polynomial in question is in fact irreducible, probably most easily checked by the last method. $\endgroup$ – B.Swan Mar 22 at 18:47

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.


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