Subfield of Reals: Q(pi^(1/3))

[ehrenfest]

[SOLVED] field theory

Homework Statement

Assume pi is transcendental over Q. Find a subfield F of the reals such that pi is algebraic of degree 3 over F.

Homework Equations

The Attempt at a Solution

Umm...the only subfield I know of the reals is the rationals. Is the answer Q(pi^(1/3))? Do people understand the that simple extension notation? How would you read Q(pi^(1/3)) in English?

 

[NateTG]

I think a field containing \pi^{\frac{1}{3}} would also contain \pi, so \pi would end up being algebraic of degree zero.

\mathbb{Q}(\sqrt[3]{\pi}) can be read as "the rational numbers adjoin the cube root of pi."

 

[ehrenfest]

I think a field containing \pi^{\frac{1}{3}} would also contain \pi, so \pi would end up being algebraic of degree zero.

Do you mean degree 1?

What should I adjoin to Q then? pi^3?

 

[masnevets]

Yes, and you use the fact that pi is transcendental to show that pi is not contained in Q(pi^3).

 

[ehrenfest]

Yes, and you use the fact that pi is transcendental to show that pi is not contained in Q(pi^3).

It just seems obvious that there is now way you get pi with a polynomial over the rationals evaluated at pi^3. Is there a better explanation?

Also, I understand why adjoining pi^(1/3) makes no sense. But can you explain the thought-process that gave away why adjoining pi^3 does give you a third-degree polynomial that has pi as a zero and why there is no first or second degree polynomial in Q(pi^3) that will have pi as a zero?

 

[NateTG]

It just seems obvious that there is now way you get pi with a polynomial over the rationals evaluated at pi^3. Is there a better explanation?

Also, I understand why adjoining pi^(1/3) makes no sense. But can you explain the thought-process that gave away why adjoining pi^3 does give you a third-degree polynomial that has pi as a zero and why there is no first or second degree polynomial in Q(pi^3) that will have pi as a zero?

The first part is easy - the polynomial with \pi as a solution will be:
x^3-\pi^3

For the second part, the existence of such polynomials would require the field to contain \pior \pi^2.