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ehrenfest

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## Homework Statement

Let E be an extension field of F, and [itex]\alpha,\beta \in E[/itex]. Suppose [itex]\alpha[/itex] is transcendental over F but algebraic over [itex]F(\beta)[/itex]. Show that \beta is algebraic over [itex]F(\alpha)[/itex].

## Homework Equations

## The Attempt at a Solution

I think it makes sense to divide into two cases:

Case 1: [itex]\beta[/itex] 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: [itex]\beta[/itex] is transcendental over F

In this case, [itex]\phi_{\beta}(F[x])[/itex] is only an integral domain, so [itex]F(\beta)[/itex] is the field of quotients of [itex]\phi_{\beta}(F[x])[/itex], call it G. Because \alpha is transcendental over F, we know that [itex]F(\alpha)[/itex] is the field of quotients of [itex]\phi_{\alpha}(F[x])[/itex], 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 [itex]p(x) = \sum_{i=0}^{\infinity}a_i x^i[/itex]. Then [itex]p_{\alpha} = \sum_{i=0}^{\infinity}a_i \alpha^i[/itex]. But what are the a_i? They are elements of G. Thus, we can rewrite p(\alpha) as

[tex] \sum_{i=0}^{\infinity}\frac{\sum_{j=0}^{\infinity}f_{ji} \beta^j}{\sum_{h=0}^{\infinity}f_{hi} \beta^h} \alpha^i[/tex]

where we know that f_ji and f_hi are in F. We know that must equal 0. That is:

[tex] \sum_{i=0}^{\infinity}\frac{\sum_{j=0}^{\infinity}f_{ji} \beta^j}{\sum_{h=0}^{\infinity}f_{hi} \beta^h} \alpha^i = 0 [/tex]

Thus we multiply both sides by [tex] \prod_{k=0}^{\infinity}\sum_{h=0}^{\infinity}f_{hk} \beta^h [/tex] to get:

[tex] \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 [/tex]

Is this getting anywhere?

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