2
$\begingroup$

I'm trying to solve the following problem.

Let $k$ be a field of characteristic $0$. Assume that for each finite extension $E$ of $k$, the index $(E^* : E^{*n})$ is finite for every positive integer n. Show that for each positive integer $n$, there exists only a finite number of abelian extensions of $k$ of degree $n$.

If $k$ contains a primitive n-th root of unity, one could use the one-to-one correspondence of abelian extension of $k$ of exponent n and subgroups of $k^*$ containing the n-th powers of the nonzero elements of $k$. For this case one of the ways to solve is as in the answer of this post: Find the bijection between Kummer's field and Galois subgroup.

But for $k$ not containing n-th roots of unity, do we have any kind of correspondence between, say, abelian extension of $k$ of exponent m and abelian extension of $k(\zeta)$ of exponent n, whence $\zeta$ is a primitive n-th root of unity?

I observed that an abelian extension of $k$ of exponent n has extension degree no more than the extension degree over $k(\zeta)$ of the abelian extension of $k(\zeta)$ of exponent n generated by the same set, multiplied by $\varphi(n)$, whence $\varphi(n)$ denotes the Euler function.

Another observation: Assume $k$ does not contain n-th roots of unity. Let H be a subgroup of $k^*$ containing the n-th powers of the nonzero elements of $k$, then $H$ and $\zeta^j$ together generates a subgroup of $k(\zeta)^*$ containing the n-th powers of the nonzero elements of $k(\zeta)$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $L/k$ be the compositum of all the abelian extensions of degree at most $n$ over $k(\zeta_n)$. Since $k$ has characteristic zero, $L/k$ is separable. Then, since $k(\zeta_n)$ has all $n$-th roots of unity, you already know that $L/k$ is finite. If $E/k$ is an abelian extension of degree $\leq n$, then $E(\zeta_n)$ is an abelian extension of $k(\zeta_n)$ of degree $\leq n$, hence $E\subset E(\zeta_n) \subset L$. Since $L/k$ is separable, it contains at most finitely many subextensions. Hence the set of possible $E$ is finite.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .