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- Jun 22, 2012

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

Determine the splitting field and its degree over [TEX] \mathbb{Q} [/TEX] for [TEX] x^4 - 2 [/TEX].

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

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

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

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

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

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

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

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

[Note that this has also been posted on MHF]

Determine the splitting field and its degree over [TEX] \mathbb{Q} [/TEX] for [TEX] x^4 - 2 [/TEX].

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

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

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

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

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

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

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

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

[Note that this has also been posted on MHF]

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