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

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

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

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

Think minimal polynomials.

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

Originally Posted by SlipEternal
Think minimal polynomials.
Thanks SlipEternal ... well, given that $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1$ the degree of minimal polynomial of the root u over $\displaystyle \mathbb{Q}(u)^2$ is 1.

Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in $\displaystyle \mathbb{Q}(u)^2$???)

How does this lead us to $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2$?

I think I need a bit more help?

Peter

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

Originally Posted by Bernhard
Thanks SlipEternal ... well, given that $\displaystyle [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1$ the degree of minimal polynomial of the root u over $\displaystyle \mathbb{Q}(u)^2$ is 1.

Thus I imagine the minimal polynomial is of the form x - u ... ... (the coefficients of the polynomial should be in $\displaystyle \mathbb{Q}(u)^2$???)

How does this lead us to $\displaystyle \mathbb{Q}(u) = \mathbb{Q}(u)^2$?

I think I need a bit more help?

Peter
That implies that $\displaystyle u \in \mathbb{Q}(u^2)$ since $\displaystyle u$ is the coefficient of $\displaystyle x^0$ in that minimal polynomial.

5. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

Originally Posted by SlipEternal
That implies that $\displaystyle u \in \mathbb{Q}(u^2)$ since $\displaystyle u$ is the coefficient of $\displaystyle x^0$ in that minimal polynomial.
Sorry SlipEternal ... still not completely following.

Let me restate where I am (thanks to your help)

We have that the degree of the minimal polynomial m(u) of $\displaystyle \mathbb{Q}(u)$ over $\displaystyle \mathbb{Q}(u^2)$ is 1.

Thus, the minimal polynomial m(u) is a monic polynomial of degree 1 with coefficients in $\displaystyle \mathbb{Q}(u^2)$ such that m(u) = 0.

So then $\displaystyle m(u) = a_0 + a_1 u = 0$ where $\displaystyle a_0, a_1 \in \mathbb{Q}(u^2)$

How do we (formally) show that $\displaystyle u \in \mathbb{Q}(u^2)$

Peter

6. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

A monic polynomial means $\displaystyle a_1 = 1$.

7. ## Re: Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

oh ... of course!

So $\displaystyle m(u) = a_0 + u = 0$ where $\displaystyle a_0 \in \mathbb{Q}(u^2)$

So $\displaystyle u = -a_0$

and $\displaystyle a_0$ and of course $\displaystyle - a_0 \in \mathbb{Q}(u^2)$

Thanks!

Peter