# How to prove that $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]=12$

I know that

$$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]=[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]\cdot[\mathbb{Q}(3^{1/4}):\mathbb{Q}],$$ and

$$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]=[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot[\mathbb{Q}(2^{1/3}):\mathbb{Q}].$$

But $$[\mathbb{Q}(3^{1/4}):\mathbb{Q}]=4$$, and $$[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$$

From here I can deduce that $$lcm(4,3)=12$$ divides $$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$$. This is, $$12 \leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$$.

I would like to show that the other inequality also holds, but I do not know how to continue here.

Another approach: I was triying to find the minimal polynomial of $$2^{1/3}$$ over $$\mathbb{Q}(3^{1/4})$$, or the minimal polynomial of $$3^{1/4}$$ over $$\mathbb{Q}(2^{1/3})$$ but I do not know how to justify that $$x^3-2$$ is irreducible over $$\mathbb{Q}(3^{1/4})$$, or that $$x^4-3$$ is irreducible over $$\mathbb{Q}(2^{1/3})$$.

Thanks for any help.

• You can assume that $x^3-2 = (x+a)(x^2+bx+c)$, where $a,b,c \in \mathbb{Q}(3^{1/4})$ and you'll find that $a^3 = -2$. Then try to show that $a \not \in \mathbb{Q}(3^{1/4})$.
– Ubik
Commented Jun 10 at 9:27
• @RHspqr If $-2^{1/3} \in \mathbb{Q}(3^{1/4})$, then $2^{1/3} \in \mathbb{Q}(3^{1/4})$. Thus, $\mathbb{Q}(2^{1/3},3^{1/4})=\mathbb{Q}(3^{1/4})$. This implies that $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=[\mathbb{Q}(3^{1/4}):\mathbb{Q}].$ Which leads to $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(2^{1/3})]\cdot3=4.$ This is, $3|4$. Contradiction. Hence, $x^3-2$ is irreducible over $\mathbb{Q}(3^{1/4})$. Finally, $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]\leq3$, giving $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}] \leq 12$. Therefore the equality Commented Jun 11 at 4:41
• @RHspqr Is this reasoning correct? Also, I would like to ask how can be proved that certain polynomial is of the least degree? for example, knowing that $x^3-2$ is irreducible over $\mathbb{Q}(3^{1/4})$, I only get that the minimal polynomial of $2^{1/3}$ over $\mathbb{Q}(3^{1/4})$ has at most degree 3. Right? Thanks! Commented Jun 11 at 4:50
• Yes that's a correct reasoning. For your question about least degree, we know that $K(\alpha) = K[x]/m_{\alpha}(x)$, which means that $m_{\alpha}(x)$ generates a maximal ideal over $K[x]$. However, since $K[x]$ is a PID, we know that all prime elements are irreducible and all prime ideals are maximal. Hence we know that $m_{\alpha} = f$.
– Ubik
Commented Jun 11 at 8:47
• We may also prove this from the definition of irreducible polynomial. By definition, the minimal polynomial divides all $f \in K[x]$ such that $f(\alpha) = 0$, thus $m_{\alpha} |f$. However, as $f$ is irreducible, $m_{\alpha} = 1 or f$.
– Ubik
Commented Jun 11 at 8:48

If $$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$$, then $$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]<12$$, against the fact that 12 divides $$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}]$$, so $$3\leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]$$.
On the other hand $$[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \leq \deg(x^3-2)= 3$$.
This works in general for $$\mathbb F(a,b)/\mathbb F$$, as soon as the degrees $$[\mathbb F(a):\mathbb F]$$ and $$[\mathbb F(b):\mathbb F]$$ are coprime (see this post).
• Then it cannot happen $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$, but $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \geq 3$ leads to $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}] \geq 12$, which is the inequality I had found. Isn't it? Commented Jun 10 at 23:56
• $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]<3$ leads to a contradiction to the inequality you proved, so $3\leq [\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})]$. On the other hand $[\mathbb{Q}(2^{1/3},3^{1/4}):\mathbb{Q}(3^{1/4})] \leq \deg(x^3-2)= 3$, and you are done. Commented Jun 11 at 8:20