Is $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$?

Question

Let $\alpha = i\sqrt[4]{2}$. Then, $1 + \alpha = 1 + i\sqrt[4]{2}$.
Then, $\mathbb{Q}(\alpha) = \{ c_0 + c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 \mid c_0, c_1, c_2, c_3 \in \mathbb{Q} \}$.
and $\mathbb{Q}(1 + a) = \{ b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 \mid b_0, b_1, b_2, b_3 \in \mathbb{Q} \}$.
Since, $b_0 + b_1 (1 + \alpha) + b_2 (1 + \alpha)^2 + b_3 (1 + \alpha)^3 = b_0 + b_1 + b_1 \alpha + b_2 + 2 b_2 \alpha + b_2 \alpha^2 + b_3 + 3 b_3 \alpha + 3 b_3 \alpha^2 + b_3 \alpha^3$
$= (b_0 + b_1 + b_2 + b_3) + (b_1 + 2b_2 + 3b_3) \alpha + (b_2 + 3b_3) \alpha^2 + b_3 \alpha^3$.
Notice that $b_0,b_1,b_2,b_3 \in \mathbb{Q}$. Then, $(b_0 + b_1 + b_2 + b_3)$, $(b_1 + 2b_2 + 3b_3)$ and $(b_2 + 3b_3)$ are all in $\mathbb{Q}$ as well.
So, let $c_0 = (b_0 + b_1 + b_2 + b_3)$, $c_1 = (b_1 + 2b_2 + 3b_3)$, $c_2 = (b_2 + 3b_3)$ and $c_3 = b_3$.
Then, $\mathbb{Q}(\alpha) = \mathbb{Q}(1 + \alpha)$.
So, $\mathbb{Q}(i\sqrt[4]{2}) = \mathbb{Q}(1 + i\sqrt[4]{2})$

Answer 1

We have $1+\alpha\in \Bbb Q(\alpha)$, since $1,\alpha\in \Bbb Q(\alpha)$, and this is a field, so that also its sum is in it. Conversely, $\alpha\pm 1\in \Bbb Q(1+\alpha)$, so that $2\alpha=(\alpha+1)+(\alpha-1)\in \Bbb Q(1+\alpha)$. Hence we have $\Bbb Q(\alpha)=\Bbb Q(1+\alpha)$.

Answer 2

Your prove is right, but remember the definition of $\mathbb{Q} (\alpha)$, the smallest field that contains $\mathbb{Q}$ and $\alpha$. So you only need to prove that $\alpha\in \mathbb{Q}(\alpha+1)$ ($\mathbb{Q}(\alpha+1)\subset\mathbb{Q}(\alpha)$) and $\alpha+1\in \mathbb{Q}(\alpha)$ ($\mathbb{Q}(\alpha) \subset\mathbb{Q}(\alpha+1)$).

This is more easy, because $1\in \mathbb{Q}$ and ${-1}\in\mathbb{Q}$.