7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Your assertion cannot be true for algebraically closed fields, because no proper algebraic extension exists in that case. $\endgroup$ – Hagen Knaf Jan 3 '14 at 2:08
  • $\begingroup$ @Hagen Good point. $\endgroup$ – Alex Becker Jan 3 '14 at 3:05
4
$\begingroup$

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 $K(f)$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – P Vanchinathan Jan 3 '14 at 14:20
4
$\begingroup$

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).
$\endgroup$
  • $\begingroup$ Thanks Hagen, for enlightening how the generalisations to higher genus fail. $\endgroup$ – P Vanchinathan Jan 3 '14 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.