Show that $\beta$ is Algebraic over $F(\alpha)$ in Simple Extensions

[ehrenfest]

Homework Statement

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 $F(\alpha)$.

Homework Equations

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?

 

[ehrenfest]

anyone?

 

[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 $\beta$.

 

[ehrenfest]

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

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?

 

[NateTG]

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?

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.

 

[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?

 

[NateTG]

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

 

[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?

 

[NateTG]

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?

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

 

[NateTG]

So, then it is a finite product. So now you think I can convolve things to get a polynomial in beta?

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