I just proved that any finite extension of fields is an algebraic extension. That is, given fields $F,K$ such that $K \subseteq F$ is a subfield then if we can view $F$ as a vector space over $K$ with finite base, then any element of $F$ is the root of a polynomial with coefficients in $K$.

I was told that the converse is not true. Do you know any example where an extension is algebraic but not finite?


marked as duplicate by M. Winter, user223391, Xander Henderson, user99914, Leucippus Oct 4 '17 at 2:41

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    $\begingroup$ The classical one: $\overline{\Bbb Q}/\Bbb Q$. $\endgroup$ – Xam Oct 3 '17 at 15:33

An algebraic extension $L/F$ is finite iff there is a positive integer $n$ such that $[K:F] \leq n$ for all finite subextensions $K$ of $L/F$:

$\bullet$ If $L/F$ is finite, then one can take $n = [L:F]$.
$\bullet$ If $[K:F] \leq n$ for all finite subextensions, let $K_0/F$ be a finite subextension with $[K_0:F]$ maximal. Then for any finite subextension $K/F$, we must have $[KK_0:F] = [K_0:F]$ so $KK_0 = K_0$ and $K \subset K_0$. Since every algebraic extension is the union of its finite subextensions, this implies $K_0 = L$.

This shows that the strategy of exhibiting finite subextensions of arbitrarily large degree done in M. Winter's answer will always work.

In fact, by a theorem of Artin-Schreier -- see e.g. Theorem 15.42 of these notes -- a field $F$ admits an infinite degree algebraic extension iff it admits a finite extension of degree at least $3$. So because the polynomial $f(x) = x^3 -2$ has no rational roots, there is an algebraic extension of $\mathbb{Q}$ of degree at least $3$ and thus an infinite degree algebraic extension of $\mathbb{Q}$. (This is a long way to go to avoid using Eisenstein's irreducibility criterion.)


Let $p$ be a prime number. The extension $\Bbb Q[\sqrt[n]{p}]/\Bbb Q$ is of order $n$. The algebraic closure $\bar{\Bbb Q}$ of $\Bbb Q$ does contain all $\sqrt[n] p$, hence cannot be of finite order.


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