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Let $F$ be a field in $K$ and $\alpha, \beta \in K$ be arbitrary (we do not assume $\alpha$ and $\beta$ are algebraic).

How do we prove that $$ F(\alpha, \beta) = \Big\{ \frac{f(\alpha,\beta)}{g(\alpha, \beta)} \big| f(\alpha,\beta), g(\alpha, \beta) \text{ are polynomial with variables $\alpha$ and $\beta$ with coefficients in } F \Big\} $$

Also if we assume $\alpha$ and $\beta$ are algebraic over $F$, can we simplify the expression to just $f(\alpha, \beta) \in F[\alpha, \beta]$?

Andy
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  • Try to prove that the RHS is a field containing $F$ , $\alpha$ and $\beta$. Can you do this from the definition of a field? Please add your attempt to the question, even if you succeed. This will certainly show that it contains $F(\alpha,\beta)$, for starters. The answer to your second question is a yes and is a reflection of the fact that the smallest ring containing a field an an element algebraic over it, is equal to the smallest field having the same property : you can look this up here. – Sarvesh Ravichandran Iyer May 27 '21 at 06:57
  • @TeresaLisbon Thanks for your comment. The proof of my first question is not too bad. But somehow, I don't feel very comfortable for this result, I guess I'm just looking for some confirmations on it. – Andy May 27 '21 at 07:05
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    Oh, you are looking for confirmation? I thought it might be better if you proved it and asked for clarifications on your proof, but I am here to confirm it for you : the result is true. The idea is what I mentioned : LHS is contained in RHS because the RHS is a field. The RHS is contained in the LHS, because if $F,\alpha,\beta$ is contained in any field, then by combining $\alpha,\beta$ etc. using addition, multiplication , subtraction and division you can create any expression of the form $\frac{f(\alpha,\beta)}{g(\alpha,\beta)}$ for any $f,g$ polynomials. – Sarvesh Ravichandran Iyer May 27 '21 at 08:09

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