# Classification of fields which are isomorphic to some finite extension

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.

• Your assertion cannot be true for algebraically closed fields, because no proper algebraic extension exists in that case. – Hagen Knaf Jan 3 '14 at 2:08
• @Hagen Good point. – Alex Becker Jan 3 '14 at 3:05

Luroth's theorem on simple transcendental extension says what you have described is possible in every degree. Instead of going upwards you just have to search inward: For any transcendental $x$ over $K$ and any polynomial $f(x)$ with coefficients in $K$ the subfield of $K(x)$ generated by $f$ over $K$ is actually isomorphic to $K(x)$. Clearly $K(x)$ is an extension of degree = degree of $f$ over $K(f)$.

• In a different direction, the classical theorem of Newton on symmetric polynomials provides examples you want: For any field $F$ the take $R=F[x_1,x_2,\ldots, x_n]$, the ring of polynomials in independent variables over $F$. Then the subring of symmetric polynomials is that generated by the elementary symmetric polynomials. Now take fraction fields. You get an extension of degre $n!$ isomorphic to itself. One can generalise to all cases where Noether's problem has positive answer. – P Vanchinathan Jan 3 '14 at 14:20

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:

1. $F$ is the function field of an elliptic curve over $K$,
2. $F$ is a rational function field (see Vanchinathan's answer).
• Thanks Hagen, for enlightening how the generalisations to higher genus fail. – P Vanchinathan Jan 3 '14 at 14:15