Splitting Fields - Dummit and Foote - Exercise 1, page 545

Dummit and Foote Exercise 1 on page 545 reads as follows:

Determine the splitting field and its degree over $\displaystyle \mathbb{Q} $ for $\displaystyle x^4 - 2 $.

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I have started on the solution to this exercise as follows:

The two roots of $\displaystyle x^4 - 2 $ are $\displaystyle \alpha = \sqrt[4]{2}$ and $\displaystyle \beta = \sqrt[4]{2}i $

Thus the splitting field is $\displaystyle \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) $

We note that $\displaystyle \mathbb{Q}(\sqrt[4]{2}, \sqrt[4]{2}i ) = \mathbb{Q}(\sqrt[4]{2}, i ) $ since the product of $\displaystyle \sqrt[4]{2} $ and i must be in $\displaystyle \mathbb{Q}(\sqrt[4]{2}, i ) $

The element $\displaystyle \alpha = \sqrt[4]{2} \in \mathbb{R} $ is algebraic over $\displaystyle \mathbb{Q} $ and so $\displaystyle m_{\alpha}(x) = m_{\sqrt[4]{2}}(x) = x^4 - 2$ is the minimal polynomial for $\displaystyle \alpha = \sqrt[4]{2}$ over $\displaystyle \mathbb{Q} $.

Thus the degree of $\displaystyle \alpha = \sqrt[4]{2}$ over $\displaystyle \mathbb{Q} $ is the degree of $\displaystyle m_{\alpha}(x) $ = 4

Hence $\displaystyle [\mathbb{Q}(\sqrt[4]{2}) \ : \ \mathbb{Q}] $ = 4

The element $\displaystyle \beta = \sqrt[4]{2}i \in \mathbb{C} $ is also algebraic over $\displaystyle \mathbb{Q} $ and so $\displaystyle m_{\beta}(x) = m_{\sqrt[4]{2}i}(x) = x^4 - 2$ is the minimal polynomial for $\displaystyle \beta = \sqrt[4]{2}i $ over $\displaystyle \mathbb{Q} $.

Thus the degree of $\displaystyle \beta = \sqrt[4]{2}i $ over $\displaystyle \mathbb{Q} $ is the degree of $\displaystyle m_{\beta}(x) $ = 4

Hence $\displaystyle [\mathbb{Q}(\sqrt[4]{2}i) \ : \ \mathbb{Q}] $ = 4

But where to from here - need to find the degree of the splitting field.

Can someone please confirm that my reasoning above is valid and show me the way forward from here?

Peter