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Let $F$ be fields. If $\beta$ is transcendental over $F$, then is that $F(\beta)$ an infinite extention of $F$ ?
Or $F(\beta)$ can be an finite extention of $F$ ?

xxxxxx
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    $F[\beta]$ is isomorphic to $F[x]$, it is an infinite dimensional $F$-vector space. $[F(\alpha):F] =n$ implies that $\alpha^n = \sum_{j=0}^{n-1} c_j \alpha^j$ for some $c_j\in F$ ie. $\alpha$ is algebraic. – reuns Feb 25 '21 at 06:53
  • @reuns this means If $\beta$ is transcendental over $F$, then $F(\beta)$ must be an infinite extention of $F$. Right ? – xxxxxx Feb 25 '21 at 07:07
  • Per reuns argument, $[F(\beta) : F]$ is finite $\implies \beta$ is algebraic. "$\beta$ is transcendental" means "$\beta$ is not algebraic". Therefore by the law of the contrapositive, "$\beta$ is trancendental" $\implies [F(\beta) : F]$ is not finite. – Paul Sinclair Feb 25 '21 at 14:38

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