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Let $K$ be a field and $L$ an extension of $K$. I wonder how much larger the multiplicative group $L^\times$ of $L$ is than the multiplicative group $K^\times$ of $K$.

I know that if $L=K(t)$ and $t$ is transcendental over $K$, then $L^\times/K^\times$ is isomorphic to the direct product of an infinite number of copies of the integers: Indeed, any (monic) rational function can be uniquely written as the product of a finite number of powers of irreducible polynomials. In particular, $L^\times/K^\times$ is not finitely generated.

What happens when $L=K(t)$ and $t$ is algebraic over $K$? I can handle the case when $K$ is finite, but what happens in the infinite case? Even for $L=Q(\sqrt{2})$ it's not immediate to me if $L^\times/K^\times$ is finitely generated or not.

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Let's look at the more familiar example $L=\mathbb{Q}(\sqrt{-1})$, $K=\mathbb{Q}$. It is easy to see that $L^\times/K^\times$ is not finitely generated. Indeed, $L^\times$ is the sum of copies of $\mathbb{Z}$ indexed by prime Gaussian integers, $K^\times$ is the sum of copies of $\mathbb{Z}$ indexed by prime integers; since infinitely many primes split in Gaussian integers, you get the result. –  t3suji Mar 16 '10 at 14:10
    
When $K=\mathbb{Q}_p$ (where $p$ is a prime) and the extension is of degree $[L:K]>1$, the group $L^\times/K^\times$ is uncountable. –  Chandan Singh Dalawat Mar 16 '10 at 14:15
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A similar, and extremely interesting, question, is the structure of $K^{\times}/N_{L/K}L^{\times}$ and its relation to $G(L\cap K^{ab}/K)$, where $L\cap K^{ab}$ is the maximal abelian extension of $K$ contained in $L$. (Probably you already know this, but I thought I'd mention it in case you didn't.) –  Joe Silverman Mar 15 '12 at 20:21
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3 Answers

up vote 12 down vote accepted

You're looking for

  • A. Brandis, Über die multiplikative Struktur von Körpererweiterungen, Math. Z. 87 (1965), 71-73

Brandis proved that $L^\times/K^\times$ is not finitely generated whenever $K$ is infinite and $L \ne K$ (thanks Pete). The claim is reduced to finite algebraic extensions of global fields, for which there are infinitely many prime ideals in $K$ that do not remain inert in $L$, and this does it.

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(And $L \neq K$.) Franz, you're fast this morning (I guess partly because it's not morning for you). You beat me to this one handily. –  Pete L. Clark Mar 16 '10 at 14:17
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I was done teaching at 13:00, had lunch, and am sipping my afternoon cop of coffee. Given that Brandis was one of my teachers, I guess it is ok that I should be the one to answer this question -) –  Franz Lemmermeyer Mar 16 '10 at 14:22
    
Perfect, this is exactly what I was looking for. Thank you! –  Guntram Mar 18 '10 at 6:51
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To add to Franz's nice answer:

Let $\mathcal{C}$ be the collection of groups isomorphic to the direct sum of a free abelian group of countable rank with a finite abelian group. In the case where $L/K$ is a nontrivial finite separable extension of global fields, each of the following groups is in $\mathcal{C}$ (and the first two are free):

1) The group of fractional ideals of $K$ (or divisors in the function field case)

2) The group of principal fractional ideals of $K$

3) $K^\times$

4) $L^\times/K^\times$

Proof: The first three can be proved in succession by using infinitude of primes, finiteness of class groups, and the Dirichlet unit theorem.

4) As suggested by t3suji and Franz, Chebotarev shows that the rank is infinite. On the other hand, the following trick shows that $L^\times/K^\times$ is a subgroup of a group in $\mathcal{C}$ (and hence in $\mathcal{C}$ itself): Replace $L$ by its Galois closure. Let $\sigma_1,\ldots,\sigma_d$ be the elements of $\operatorname{Gal}(L/K)$. Then $$x \mapsto (\sigma_1(x)/x,\ldots,\sigma_d(x)/x)$$ injects $L^\times/K^\times$ into $L^\times \times \cdots \times L^\times$, which is in $\mathcal{C}$.

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Perhaps useful to you: As a replacement of finitly generated in a local situation, one can sometimes use that the quotient is cocompact.

For a local field, we have in the usual scaling of the multiplicative Haar measure that the quotient measure $F^\times / K^\times$ is the ramification index.

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