TITLE: Classification of fields which are isomorphic to some finite extension
QUESTION [7 upvotes]: For what fields $K$ do there exist finite extensions $E/K$ (of degree $>1$) such that $K\cong E$?
If $K$ has finite degree over its prime field, then any finite extension $E/K$ has greater degree over the prime field, therefore $E\not\cong K$.
On the other hand, if $K=k(x)$ for some field $k$, then adjoining a formal square root $y$ of $x$ gives an extension $K(y)/K$ of degree $2$ such that $K\cong K(y)$, since $K=k(x)\cong k(y)=K(y)$.
I suspect the same holds for any field which contains an element which is transcendental over its prime subfield. However, I'm at a loss for how to deal with cases like infinite algebraic extensions.

REPLY [4 votes]: In contrast to Vanchinathan's answer there are many fields, that do not have the property under discussion: consider the function field $F$ of an algebraic curve over a field $K$, or in other words a finitely generated field extension $K\subset F$ of transcendence degree $1$. Assume that $K$ is the algebraic closure of a finite field or of the rationals to make the connection to the original post of Axel Becker. Let $E\supset F$ be a finite separable extension of degree $n>1$ isomorphic to $F$ via $\sigma$ say. Then $\sigma$ leaves the elements of $K$ fix and we can apply the genus formula of Riemann-Hurwitz to get:
$2g_E-2=n(2g_F-2)+d$,
where $d\geq 0$. Since $F$ and $E$ are ismorphic over $K$ we get $g_F=g_E$ and we see that the equation can never be true in the case $g_F>1$. The remaining cases are the following:

$F$ is the function field of an elliptic curve over $K$,
$F$ is a rational function field (see Vanchinathan's answer).