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

• Oct 2nd 2013, 01:35 AM
Bernhard
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