# Simple extensions

1. Feb 14, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Let E be an extension field of F, and $\alpha,\beta \in E$. Suppose $\alpha$ is transcendental over F but algebraic over $F(\beta)$. Show that \beta is algebraic over $F(\alpha)$.

2. Relevant equations

3. The attempt at a solution
I think it makes sense to divide into two cases:
Case 1: $\beta$ is algebraic over F
It is obvious that if \beta is algebraic over F then, \beta will be algebraic over any extension field of F, right?
Case 2: $\beta$ is transcendental over F
In this case, $\phi_{\beta}(F[x])$ is only an integral domain, so $F(\beta)$ is the field of quotients of $\phi_{\beta}(F[x])$, call it G. Because \alpha is transcendental over F, we know that $F(\alpha)$ is the field of quotients of $\phi_{\alpha}(F[x])$, call it H. It is obvious that G and H are subfields of the field E. We know that there is an irreducible polynomial p(x) in G that has \alpha as a zero. But we want a polynomial in H[x] that has \beta as a zero.
Say $p(x) = \sum_{i=0}^{\infinity}a_i x^i$. Then $p_{\alpha} = \sum_{i=0}^{\infinity}a_i \alpha^i$. But what are the a_i? They are elements of G. Thus, we can rewrite p(\alpha) as
$$\sum_{i=0}^{\infinity}\frac{\sum_{j=0}^{\infinity}f_{ji} \beta^j}{\sum_{h=0}^{\infinity}f_{hi} \beta^h} \alpha^i$$
where we know that f_ji and f_hi are in F. We know that must equal 0. That is:
$$\sum_{i=0}^{\infinity}\frac{\sum_{j=0}^{\infinity}f_{ji} \beta^j}{\sum_{h=0}^{\infinity}f_{hi} \beta^h} \alpha^i = 0$$
Thus we multiply both sides by $$\prod_{k=0}^{\infinity}\sum_{h=0}^{\infinity}f_{hk} \beta^h$$ to get:
$$\sum_{i=0}^{\infinity}\left( \prod_{k\neq i}^{\infinity}\sum_{h=0}^{\infinity}f_{hk} \beta^h \right) \sum_{j=0}^{\infinity} f_{ji} \beta^j \alpha^i = 0$$

Is this getting anywhere?

Last edited: Feb 14, 2008
2. Feb 14, 2008

### ehrenfest

anyone?

3. Feb 14, 2008

### NateTG

I think that if $\alpha$ is transcendental in $F$, and algebraic in $F(\beta)$, $\beta$ is going to be transcendental in $F$ as well.

You're close to the finish line. Simply regroup so that you've got things in order of powers of $\beta$.

4. Feb 14, 2008

### ehrenfest

If beta is algebraic in F, and alpha is algebraic in F(\beta), then we have a polynomial p(x) in F(\beta)[x] that has alpha as a zero. Thus,

$$p(\alpha) = \sum_{i=0}^{\infty}\left(\sum_{j=0}^{\infty}f_{ji}\beta^j \right)\alpha^i = 0$$

Why does that imply that \alpha is algebraic over F?

5. Feb 14, 2008

### NateTG

Leaving the line in the proof is fine, but I think if $\beta$ is algebraic in $F$ then it should be possible to multiply any polynomial in $F(\beta)$ by conjugates to get a polynomial in $F$ since $\beta$ can be written as an expression involving roots in $F$.

I assume you manged to finish the proof.

6. Feb 14, 2008

### ehrenfest

No. I did not finish the proof. So, I need to convolve an infinite product of formal polynomials, right? I can convolve two, but I have no idea how to convolve an infinity number of them. Is multiplication by an infinite number of polynomials even defined for formal power series? Don't you get infinite coefficients?

Last edited: Feb 15, 2008
7. Feb 15, 2008

### NateTG

Sorry, I missed that in the first post. Are you sure the sums should be infinite?

8. Feb 15, 2008

### ehrenfest

Well, the sum over i is a formal power series. But we know that only finitely many of the coefficients of powers of alpha are nonzero. So, I guess you could write it as:

$$\sum_{i_n}^{\infinity}\frac{\sum_{j=0}^{\infinity} f_{ji_n} \beta^j}{\sum_{h=0}^{\infinity}f_{hi_n} \beta^h} \alpha^{i_n} = 0$$

where n is the index set of the i's that index nonzero quotients of polynomials. So, then

$$\sum_{i_n=0}^{\infinity}\left( \prod_{k_m\neq i_n}^{\infinity}\sum_{h=0}^{\infinity}f_{hk_m} \beta^h \right) \sum_{j=0}^{\infinity} f_{ji_n} \beta^j \alpha^{i_n} = 0$$

where m is the same index set as n.
So, then it is a finite product. So now you think I can convolve things to get a polynomial in beta?

9. Feb 15, 2008

### NateTG

Unless the sums in the quotient are finite, you can't get a polynomial of finite degree. (You'll get a power series instead.)

10. Feb 19, 2008

### NateTG

No. You'll get $\beta^\infty$ unless the sums in the quotient are finite.