# Field theory

1. Feb 13, 2008

### ehrenfest

[SOLVED] field theory

1. The problem statement, all variables and given/known data
Assume pi is transcendental over Q. Find a subfield F of the reals such that pi is algebraic of degree 3 over F.

2. Relevant equations

3. 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?

Last edited: Feb 13, 2008
2. Feb 13, 2008

### 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."

3. Feb 13, 2008

### ehrenfest

Do you mean degree 1?

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

4. Feb 13, 2008

### masnevets

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

5. Feb 13, 2008

### ehrenfest

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?

6. Feb 14, 2008

### NateTG

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

For the second part, the existance of such polynomials would require the field to contain $\pi[itex] or [itex]\pi^2$.