Example 13 from Nicholson: Introduction to Abstract Algebra, Section 6.2, page 282 reads as follows: (see attachment)

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

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The solution comes down to the following:

Given $\displaystyle \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q} $

so $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] $

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

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

Then Nicholson (I think) concludes that $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2 $

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My problem is as follows:

How (exactly) does it follow that:

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

Can someone help?

Peter