Minimal polynomial of $\sqrt[3]{2} + \sqrt{3}$

Question

Suppose I want to find the minimal polynomial of the number $\sqrt[3]{2} + \sqrt{3}$.

Now that means I want to find a unique polynomial that is irreducible over $\Bbb Q$ such that $f(x)=0$. Now I know that because $\sqrt[3]{2} + \sqrt{3}$ belongs to $\Bbb Q( \sqrt[3]{2} , \sqrt{3})$ it might be degree $2$, $3$ or $6$ and does not belong to $\Bbb Q( \sqrt[3]{2})$ so it cannot be of degree $3$ or $\Bbb Q( \sqrt{3})$ so it cannot be of degree $2$.

So it is of degree $6$. I think my sayings are a bit intuitive and not formal and lack rigorous. Couldn't it belong to another extension of degree $2$? or $3$? I can't answer that. Why checking only those $2$ is enough? Or is it wrong at all to say that?

Answer 1

You are correct that $\def\Q{\Bbb Q}x=2^{1/3}+3^{1/2}\in \Q( \sqrt[3]{2} , \sqrt{3})$. You are also correct that the degree of the minimal polynomial $f_x$ will equal the extension degree $[\Q(x):\Q]$, and hence $\deg f_x\mid [\Q( \sqrt[3]{2} , \sqrt{3}):\Q]=6$. However, there might be more intermediate fields $\Q( \sqrt[3]{2} , \sqrt{3})\supset \Q(\alpha)\supset \Bbb Q$ then just $\Q(2^{1/3})$ and $\Q(3^{1/2})$. In particular, $\Q(x)$ might just be a third field that is strictly between $\Q( \sqrt[3]{2} , \sqrt{3})$ and $\Q$. Hence your reasoning in incomplete.

One can show however, that $\Q(3^{1/2})\subset \Q(x)$ and $\Q(2^{1/3})\subset \Q(x)$. This would force the degree $[\Q(x):\Q]$ to be six, since now $$[\Q(x):\Q]=[\Q(x):\Q(3^{1/2})][\Q(3^{1/2}):\Q]=2[\Q(x):\Q(3^{1/2})]$$ $$[\Q(x):\Q]=[\Q(x):\Q(2^{1/3})][\Q(2^{1/3}):\Q]=3[\Q(x):\Q(2^{1/3})]$$ Hence $2\mid [\Q(x):\Q]$ and $3\mid [\Q(x):\Q]$. Also, we already saw $[\Q(x):\Q]\leq [\Q(3^{1/2},2^{1/3}):\Q]=6$. Hence we conclude that $[\Q(x):\Q]=6$.

Answer 2

Let $a=\sqrt[3]{2} + \sqrt{3}$. Notice that $$(a-\sqrt{3})^3=2=a^3-3\sqrt 3 a^2+9a-3\sqrt 3 = a^3+9a-\sqrt 3 (3a^2+3) \tag 1$$

therefore

$$\sqrt 3 = \frac{a^3+9a-2}{3a^2+3} \tag 2$$

In particular, $\Bbb Q(a)$ contains $\Bbb Q(\sqrt 3)$ and also contains $\Bbb Q(a-\sqrt 3) = \Bbb Q(\sqrt[3]{2})$. Therefore your intuition is correct: the degree of $\Bbb Q(a)$ is a multiple of $3$ and a multiple of $2$ (over $\Bbb Q$).

The degree of the minimal polynomial of $a$ over $\Bbb Q$ is then at least $6$.

From $$(a^3+9a-2)^2 = [\sqrt 3 (3a^2+3)]^2 \tag 3$$ you get a monic polynomial $P \in \Bbb Q[X]$ of degree $6$, such that $P(a)=0$. Thus $P$ is the minimal polynomial of $a$ over $\Bbb Q$.

Here is the minimal polynomial $P(X)$ of $a$ over $\Bbb Q$ :

$P(x) = x^6-9 x^4-4 x^3+27 x^2-36 x-23$.

Answer 3

The number requires a sixth-order polynomial to solve. The intuition here is that for belonging to $Z(\sqrt{3}, \sqrt[3]{2})$ that it actually belongs to say $Z(1,\sqrt{3})(1,\sqrt[3]{2},\sqrt[3]{4})$.

One might note that these pair of systems are closed to multiplication, addition and subtraction, and that the appending of the $(1,\sqrt{3})$ will either result in an integer, or a number oin this set. Since the intersection of this set and the second set $(1,\sqrt[3]{2},\sqrt[3]{4})$ is the integers only (ie $Z$), then because the solution is in neither of the axial sets, it is in the union.

This particular set does not provide any means for a subset other than the two axial sets, in that no subset of the six base units make a set closed to multiplication. And there are six units, and so it must solve a calculation in terms of a sixth order equation.

An example where one might get a possible subset is $Z(1,\sqrt[3]{2},\sqrt[3]{4})(1,\sqrt[3]{2},\sqrt[3]{4})$ which contains possible sets like $Z(1,\sqrt[3]{2},\sqrt[3]{36})$ and $Z(1,\sqrt[3]{12},\sqrt[3]{18})$.