It is known that an extension generated by a finite number of algebraic elements is a finite extension. Is it also true if the elements are not algebraic?

Or in other words, give an example of a finitely generated field extension that is not algebraic.

  • $\begingroup$ No. A finite extension is always algebraic. $\endgroup$ – Jyrki Lahtonen Nov 22 '15 at 11:44
  • $\begingroup$ What would be a relevant example to show that this does not hold? $\endgroup$ – user240035 Nov 22 '15 at 11:46
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    $\begingroup$ $F(t)$ is finitely generated (as a field extension), but not a finite extension, if $t$ is an indeterminate. $\endgroup$ – Gerry Myerson Nov 22 '15 at 11:53
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    $\begingroup$ $\Bbb{Q}(\pi)$ is finitely generated, but not finite. For if it were finite, then $\pi$ would be algebraic, which it is not (though the proof is too deep to be covered here). This is a special case of Gerry Myerson's umbrella answer. Only this with a specific transcendental in place of $t$, and $\Bbb{Q}$ in place of $F$. All this while also obfuscating the relevant points. I hope you understand why I won't post this as an answer. $\endgroup$ – Jyrki Lahtonen Nov 22 '15 at 12:02
  • $\begingroup$ To elaborate a bit on Jyrki's comment, it was long unknown whether there existed any non-algebraic elements in $\Bbb R$ (which is obviously an extension field of $\Bbb Q)$. This is a big problem, since $\Bbb R$ is uncountable, and the algebraic reals are countable (thus, most real numbers ought to be transcendental). Liouville proved that Liouville's constant $\sum_k 10^{-k!}$ was transcendental in 1851, and Hermite proved $e$ was transcendental in 1873 (both these precede Cantor's uncountablility proofs of 1874). The theorem Jyrki refers to is the Lindemann-Weierstrass theorem (1885). $\endgroup$ – David Wheeler Nov 22 '15 at 14:05

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