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

Question

I was hoping that you all could enlighten me as to the different methods of showing that $[\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/2})]=3$. I know there are a few different ways of seeing this, and I would appreciate y'all showing me a few. Thanks!!

Answer 1

Method 1: Recall that $[L:K][K:F] = [L:F]$, note that $\mathbb{Q}(2^{1/3}, 2^{1/2}) = \mathbb{Q}(2^{1/6})$, and show that $X^6 - 2$ is irreducible over $\mathbb{Q}$ using Eisenstein or similar.

Method 2: If $X^3 - 2$ is reducible over $\mathbb{Q}(2^{1/2})$, then it must have a linear factor $X - \alpha$ where $\alpha = a + b \sqrt{2} \in \mathbb{Q}(2^{1/2})$ satisfies $\alpha^3 = 2$. Show that no such $\alpha$ can exist.

Answer 2

The easiest way that I see this is the standard way of using the multiplicative property of dimension of field extension. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$, and $[\mathbb{Q}(2^{1/2}):\mathbb{Q}]=2$. Now, $[\mathbb{Q}(2^{1/3}, 2^{1/2}):\mathbb{Q}]= [\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/2})].[\mathbb{Q}(2^{1/2}):\mathbb{Q}]$. Also, $[\mathbb{Q}(2^{1/3}, 2^{1/2}):\mathbb{Q}]= [\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/3})].[\mathbb{Q}(2^{1/3}):\mathbb{Q}]$

Now, since $\deg_{\mathbb{Q}}(2^{1/3}) = 3$, hence $\deg_{\mathbb{Q}(2^{1/2})}(2^{1/3}) \leq 3$. Similarly, $\deg_{\mathbb{Q}(2^{1/3})}(2^{1/2}) \leq 2$. Let $\deg_{\mathbb{Q}(2^{1/2})}(2^{1/3}) = p$, and $\deg_{\mathbb{Q}(2^{1/3})}(2^{1/2}) = q$. Note that therefore $p = [\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/2})]$ and $q = [\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/3})]$.

So, we have that $[\mathbb{Q}(2^{1/3}, 2^{1/2}):\mathbb{Q}] = 2p = 3q$, $p \leq 3$ and $q \leq 2$. Since $2$ and $3$ are coprime, and $3 \mid 2p \implies 3 \mid p$ and $p \leq 3 \implies p = 3$. Similarly, $q = 2$. Hence, $\deg_{\mathbb{Q}(2^{1/3})}(2^{1/2}) = 3$. So, $[\mathbb{Q}(2^{1/3},2^{1/2}):\mathbb{Q}(2^{1/2})]=3$.

Answer 3

Let $F_1=Q(2^{1/2}), F_2=Q(2^{1/3}), F_3=F_1(2^{1/3})$, then dimension of $F_1, F_2$ over $Q$ are 2, 3, respectively. Hence the dimension $d$ of $F_3$ over $F_1$ is at most 3.

Certainly $d$ can not be 1 since otherwise $2^{1/3}$ (which has degree 3) would be in $F_2$ an extension of dimension 2.

If $d=2$ then the dimension of $F_3$ over $Q$ is 4; since $F_2$ belongs to $F_3$ the dimension of $F_3$ is divisible by 3.

Thus $d=3$.