(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let E be an extension field of a field F. Given [tex]\alpha\in E[/tex], show that, if [tex]\alpha\notin F_E[/tex], then [tex]\alpha[/tex] is transcendental over [tex]F_E[/tex].

2. Relevant equations

[tex]F_E[/tex] denotes the algebraic closure of F in its extension field E.

3. The attempt at a solution

First, I assumed [tex]\alpha \in E[/tex] was algebraic over [tex]F_E[/tex] and that [tex]\alpha \notin F_E[/tex] and attempted to derive a contradiction. If [tex]\alpha[/tex] is algebraic over E, then [tex]F_E(\alpha)[/tex] is a finite (and therefore algebraic) extension of [tex]F_E[/tex]. Hence, [tex]\left\{1, \alpha^1,\alpha^2, ... , \alpha^{n-1} \right\}[/tex] is a basis over [tex]F_E(\alpha)[/tex] given deg([tex]\alpha, F_E[/tex]) = n, for n a natural number.

I then tried to prove that, since [tex] \left\{ 1, \alpha^1,\alpha^2, ... , \alpha^{n-1}, \alpha^n \right\} [/tex] is linearly dependent over [tex]F_E[/tex] (since it contains (n+1) elements), it must also be linearly dependent over F.

Upon proving this, my strategy was to conclude that, if [tex]\left\{ 1, \alpha^1, \alpha^2, ..., \alpha^{n-1}, \alpha^n \right\}[/tex] is linearly dependent over F, then [tex]\alpha[/tex] is a zero of some polynomial [tex]f(x) \in F[x][/tex] and hence is algebraic over F[x]. Thus, [tex] \alpha \in F_E [/tex], since [tex]\alpha \in E[/tex].

I made another attempt at a separate solution, if you'd like to see it.

Any prodding in the right direction would be much appreciated.

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# Homework Help: Transcendental over an algebraic closure of F in E

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