Find a primitive element for the extension $Q(3^{1/4}, i)/Q$ Find a primitive element for the extension $Q(3^{1/4}, i)/Q$
So, I was guessing the primitive element is $3^{1/4}+i$, and I don't have any trouble to show that $Q(3^{1/4}+i)$ is subset of $Q(3^{1/4}, i)$.
Next, we need to show $Q(3^{1/4}, i)$ is subset of $Q(3^{1/4}+i)$, which means we need to show $3^{1/4}\in{Q(3^{1/4}+i)}$ and $i\in{Q(3^{1/4}+i)}$.
I have no clue how to do it. Can any one help me on this question?
 A: By Eisenstein you can show $(x-i)^4-3$ is minimal polynomial of $3^{1/4}+i$ $\mathbb{Q}[i]$. And so, $[\mathbb{Q}(3^{1/4}+i):\mathbb{Q}(i)]=4$. So, as $[\mathbb{Q}(i):\mathbb{Q}]=2$ we've $[\mathbb{Q}(3^{1/4}+i):\mathbb{Q}]=8$ and so you get what you want
A: Old question, but I just had to deal with it. I presume there also is an elegant solution, but I haven’t found one so far.
One’s strategies are to find some $α ∈ ℚ(\sqrt[4] 3, \mathrm i)$ and then do one of the following:

*

*Show that $\sqrt[4] 3, \mathrm i ∈ ℚ(α)$.

*Compute that $[ℚ(\sqrt[4] 3, \mathrm i) : ℚ] = [ℚ(α) : ℚ]$.

One can also do a combination of both. Often, no precise computation is needed, but a convergence of general observations may suffice. For instance, to find the primitive element of a simple extension $L / K$, one could pick an element $α ∈ L$ such that $K(α) ≠ K(α^2)$ and $[K(α^2): K] ≥ 3$, but $[L : K] \mid 8$ , which already entails $L = K(α)$.
Here, one can

*

*pick $α = \sqrt[4] 3 + \mathrm i$ and $β = \sqrt[4] 3 - \mathrm i$,

*note that $α + β = 2\sqrt[4] 3$ and $α - β = 2\mathrm i$,

*conclude that $ℚ(\sqrt[4] 3, \mathrm i) = ℚ(α, β)$,

*observe that $α^2 = \sqrt 3 + 2\sqrt[4] 3 \mathrm i - 1 ∈ ℚ(\sqrt[4] 3 \mathrm i)$, so $ℚ(α^2) ⊆ ℚ(\sqrt[4] 3 \mathrm i)$,

*note that the only proper intermediate field of $ℚ(\sqrt[4] 3\mathrm i) / ℚ$ is $ℚ(\sqrt 3)$,

*to find that $\sqrt 3 ∈ ℚ(α^2) ⊆ ℚ(α)$,

*to finally conclude that $ℚ(α) = ℚ(α, β) = ℚ(\sqrt[4] 3, \mathrm i)$ since $αβ = \sqrt 3 + 1$.

To show that $ℚ(\sqrt[4] 3\mathrm i) / ℚ$ has exactly one proper subfield, observe that $ℚ(\sqrt[4] 3\mathrm i) \cong ℚ(\sqrt[4] 3)$, and since $[ℚ(\sqrt[4] 3) : ℚ] = 4$, any proper subfield $F$ of this extension has $[ℚ(\sqrt[4] 3) : F] = 2$ and, splitting $X^4 - 3$, one can see that the only real divisor of $X^4 - 3$ of degree $2$ which has $\sqrt[4] 3$ as a root is $X^2 - \sqrt 3$, so it’s the only possible candidate for aminimal polynomial of $\sqrt[4] 3$ over $F$, proving $\sqrt 3 ∈ F$, hence $F = ℚ(\sqrt 3)$.
