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<math>2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}}</math> | <math>2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}}</math> | ||
| − | So <math>\mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha)</math>. | + | So <math>\mathbb{Q}(2^{\frac{1}{3}}) \subset \mathbb{Q}(\alpha)</math> and <math>\mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha)</math>. |
Thus <math>3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}|</math>. | Thus <math>3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}|</math>. | ||
Latest revision as of 04:11, 3 July 2013
Problem 5
Clearly $ \mathbb{Q}(\alpha)\subset \mathbb{Q}(2^{\frac{1}{3}}) $. But also,
$ 2\alpha - \alpha = 16-17 \cdot 2^{\frac{1}{3}} $
So $ \mathbb{Q}(2^{\frac{1}{3}}) \subset \mathbb{Q}(\alpha) $ and $ \mathbb{Q}(2^{\frac{1}{3}}) = \mathbb{Q}(\alpha) $.
Thus $ 3=|\mathbb{Q}(2^{\frac{1}{3}}):\mathbb{Q}| = |\mathbb{Q}(\alpha):\mathbb{Q}| $.
