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I'm a self learning non-student, and I was working on some field theory from Fraleigh's text. There is a question which is numbered (29) in the document linked below. I cannot understand the solution. My question is how it is known that $F(\beta) = ${$f(\beta)/g(\beta) \, | \, where \, f(x), g(x) \in F[x]$}. My understanding was that if an element $\alpha \in E$ was transcendental then we could apply that any element in F($\alpha$) was of the form $f(\alpha)/g(\alpha)$. Here it is not given that $\beta$ is transcendental over F, but it is $\alpha$ which is transcendental over F.

Link: http://www.ms.uky.edu/~okeefea/Teaching_files/Sect%2029%20Solutions.pdf

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    This is the definition of the notation $F(\beta)$ it is rational functions in $\beta$ with coefficients in $F$. So it is not known "because" of anything: it is the definition. – 2017-02-06
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    @ Adam Hughes I don't see then that they use that alpha is transcendental over F at all anywhere in the proof. – 2017-02-06
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    I'm not sure I understand your response. I'm just noting that $F(\beta)$ or $F(\alpha)$ or whatever is just rational functions in the thing(s) inside the parentheses. How this definition is used in a proof is a wholly separate issue from what the notation means. – 2017-02-06
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    Adam : it's not really the definition, the correct answer is that of arctic tern – 2017-02-06

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If $K/F$ is a field extension and $\beta\in K$ then $F(\beta)$ is the "smallest" subfield of $K$ that contains both the subfield $F$ and the element $\beta$. Among all subfields of $K$ with this property, is is the unique minimal one (where such subfields are partially ordered by containment), and it also equals the intersection of all of the subfields with this property (containing both $K$ and $\beta$).

Consider the subset $S$ of $K$ consisting of all elements expressible as rational expressions in $\beta$ with coefficients in $F$. This is closed under addition, subtraction, multiplication and division - so it is a subfield of $K$. Moreoever, it contains $F$ and $\beta$. Conversely, suppose $L$ is any subfield of $K$ that contains both $F$ and $\beta$. Then since $L$ is closed under the field operations, $L$ must also contain every rational expression in $\beta$ with coefficients in $F$, in other words $S\subseteq L$. Therefore, $S$ is the minimal subfield with this property, and $S=F(\beta)$.