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

• ehrenfest
I was just pointing out that the proof is not finished.In summary, the conversation discusses a proof involving an extension field and two elements, one transcendental and one algebraic. The proof is divided into two cases and involves convolving an infinite product of formal polynomials. However, it is pointed out that the proof is not finished as the sums in the quotient need to be finite in order to obtain a polynomial of finite degree.
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 $F(\alpha)$.

## 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:
anyone?

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 $\beta$.

NateTG said:
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?

ehrenfest said:
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.

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

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?

ehrenfest said:
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.)

ehrenfest said:
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.

## What does it mean for $\beta$ to be algebraic over $F(\alpha)$?

For $\beta$ to be algebraic over $F(\alpha)$ means that there exists a polynomial $f(x) \in F(\alpha)[x]$ such that $f(\beta) = 0$. In other words, $\beta$ is a root of a polynomial with coefficients in the field $F(\alpha)$.

## What is a simple extension?

A simple extension is a field extension $E/F$ where $E = F(\alpha)$ for some element $\alpha \in E$. In other words, $E$ is generated by adding a single element $\alpha$ to the field $F$.

## How do you show that $\beta$ is algebraic over $F(\alpha)$?

To show that $\beta$ is algebraic over $F(\alpha)$, we need to find a polynomial $f(x) \in F(\alpha)[x]$ such that $f(\beta) = 0$. This can be done by constructing a polynomial with $\beta$ as a root or by showing that $\beta$ satisfies a polynomial equation with coefficients in $F(\alpha)$.

## What is the degree of $\beta$ over $F(\alpha)$?

The degree of $\beta$ over $F(\alpha)$ is the degree of the minimal polynomial of $\beta$ over $F(\alpha)$. In other words, it is the smallest degree of a polynomial in $F(\alpha)[x]$ that has $\beta$ as a root.

## Why is showing that $\beta$ is algebraic over $F(\alpha)$ important?

Showing that $\beta$ is algebraic over $F(\alpha)$ is important because it allows us to extend the field $F(\alpha)$ to include $\beta$. This is useful in many algebraic constructions and can help us solve problems in abstract algebra and number theory.

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