- Thread starter
- #1

- Jun 22, 2012

- 2,918

------------------------------------------------------------------------------------------------------------------

Example 13: If [TEX] u = \sqrt[3]{2} [/TEX] show that [TEX] \mathbb{Q}(u) = \mathbb{Q}(u)^2 [/TEX]

------------------------------------------------------------------------------------------------------------------

The solution comes down to the following:

Given [TEX] \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q} [/TEX]

so [TEX] [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] [/TEX]

Now Nicholson shows that [TEX] [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = 3 [/TEX] and [TEX] [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] = 3 [/TEX]

so [TEX] [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 [/TEX]

Then Nicholson (I think) concludes that [TEX] \mathbb{Q}(u) = \mathbb{Q}(u)^2 [/TEX]

----------------------------------------------------------------------------------------------------------

My problem is as follows:

How (exactly) does it follow that:

[TEX] [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 \Longrightarrow \mathbb{Q}(u) = \mathbb{Q}(u)^2 [/TEX]

Can someone help?

Peter