Finding degree and basis for a field extension

Question

I want to find the degree and basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ over $\mathbb{Q}(\sqrt{2})$. I can show that the degree is $3$, since $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}] = 6$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Intuitively, I think $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ should be a basis, which means I should show that $x^3 - 3$ is irreducible over $\mathbb{Q}(\sqrt{2})$.

How do I do that?

Answer 1

Consider a typical diamond of fields. First determine $\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})$ over the rationals. Verify that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$. This is since $x^3-3$ is irreducible by Eisenstein with $p=3$ and $x^2-2$ irreducible over the rationals as well thus,

$$\text{gcd}([\mathbb{Q}(\sqrt{2}):\mathbb{Q}],[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}])=1 \Rightarrow [\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$$ Thus if you consider $\mathbb{Q}\subset \mathbb{Q}(\sqrt{2}) \subset\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ then we must have $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}:\mathbb{Q}(\sqrt{2})]=3$ since we have that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$; hence just apply the tower theorem.

A vector space basis is $\{1,\sqrt[3]{3},\sqrt[3]{9}\}$ so your intitution was correct.

Answer 2

Your intuition is correct.

A degree three polynomial is reducible if and only if it has a root. Suppose $x^3-3$ has a root $a+b\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$; then $$ a^3+3a^2b\sqrt{2}+6ab^2+2b^3\sqrt{2}-3=0 $$ implies $$ \begin{cases} a^3+6ab^2=3 \\[4px] 3a^2b+2b^3=0 \end{cases} $$ The second equation yields $b=0$ or $3a^2+2b^2=0$ and it's easy to conclude the root does not exist.

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A proof based on dimensions is possible, too. We have $$ n=[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}),\mathbb{Q}]\tag{*} $$ so $2\mid n$; similarly $$ n=[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt[3]{3})] [\mathbb{Q}(\sqrt[3]{3}),\mathbb{Q}]\tag{**} $$ so $3\mid n$. Hence $6\mid n$. On the other hand, the two relation above show also $n\le 6$, hence $n=6$. Then (*) proves $$ [\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2})]=3 $$ and, since $x^3-3$ has $\sqrt[3]{3}$ as root, it must be irreducible over $\mathbb{Q}(\sqrt{2})$: indeed it divides the minimal polynomial of $\sqrt[3]{3}$, which must have degree $3$.