# Counter-example: any algebraic extension is finite [duplicate]

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, LeucippusOct 4 '17 at 2:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

## 2 Answers

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.