# Thread: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

1. ## Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions.

Example 13 on page 282 (see attachment) reads as follows:

"If $u = \sqrt[3]{2}$ show that $\mathbb{Q}(u) = \mathbb{Q}(u^2)$"

In the third line of the explanation - see page 282 of attachment - we read:

"But $[\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1$ because $u^2 \notin \mathbb{Q}$ ... ... "

Can someone explain why it follows that $u^2 \notin \mathbb{Q} \Longrightarrow [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1$

Peter

2. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

$[\mathbb{Q}(k) : \mathbb{Q}] = 1$ implies that the minimal polynomial for $k$ over $\mathbb{Q}[x]$ has degree 1. Hence the minimal polynomial is $x-k$ which implies $k \in \mathbb{Q}$.

3. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

Thanks SlipEternal

Just one further point of clarification:

I can see why the minimal polynomial is x - k , but why exactly does that imply that $k \in \mathbb{Q}$?

Peter

4. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page

Originally Posted by Bernhard
Thanks SlipEternal

Just one further point of clarification:

I can see why the minimal polynomial is x - k , but why exactly does that imply that $k \in \mathbb{Q}$?

Peter
Since $(x-k) \in \mathbb{Q}[x]$, that means that the coefficients of each term are elements of $\mathbb{Q}$. Hence $-k \in \mathbb{Q}$, so $k \in \mathbb{Q}$.