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

Solution 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 $[\Q(x):\Q]=6$.

Solution 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 $\Bbb Q$ :