What's the analogue of the Hilbert class field in the following analogy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:44:09Z http://mathoverflow.net/feeds/question/10947 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy What's the analogue of the Hilbert class field in the following analogy? Qiaochu Yuan 2010-01-06T19:26:07Z 2010-01-06T22:50:27Z <p>There's a wonderful analogy I've been trying to understand which asserts that field extensions are analogous to covering spaces, Galois groups are analogous to deck transformation groups, and algebraic closures are analogous to universal covering spaces, hence the absolute Galois group is analogous to the fundamental group. (My vague understanding is that the machinery around etale cohomology makes this analogy precise.) </p> <p>Does the Hilbert class field (of a number field) fit anywhere into this analogy, and how?</p> <p>Phrased another way, what does the Hilbert class field of the function field of a nonsinguar curve defined over $\mathbb{C}$ (say) look like geometrically? </p> http://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy/10950#10950 Answer by S. Carnahan for What's the analogue of the Hilbert class field in the following analogy? S. Carnahan 2010-01-06T19:48:12Z 2010-01-06T19:48:12Z <p>I guess you're looking for the maximal unramified abelian extension. The fundamental group of a genus g curve has abelianization isomorphic to $\mathbb{Z}^{2g}$, so if g is positive, you get an infinite extension. The corresponding cover is not an algebraic curve in the usual sense of the word, although I suppose you can write down a scheme of infinite type.</p> <p>Perhaps a better answer is: questions of this type fall into the domain of "geometric class field theory". If I'm not mistaken, Serre's <i>Algebraic groups and class fields</i> covers a lot the ideas.</p> http://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy/10952#10952 Answer by Pete L. Clark for What's the analogue of the Hilbert class field in the following analogy? Pete L. Clark 2010-01-06T19:55:39Z 2010-01-06T22:50:27Z <p>This is a great question. Someone will come along with a better answer I'm sure, but here's a bit off the top of my head:</p> <p>1) The Hilbert class field of a number field $K$ is the maximal everywhere unramified abelian extension of $K$. (Here when we say "$K$" we really mean "$\mathbb{Z}_K$", the ring of integers. That's important in the language of etale maps, because any finite separable field extension is etale.)</p> <p>In the case of a curve over $\mathbb{C}$, the "problem" is that there are infinitely many unramified abelian extensions. Indeed, Galois group of such is the abelianization of the fundamental group, which is free abelian of rank $2g$ ($g$ = genus of the curve). Let me call this group G. </p> <p>This implies that the covering space of C corresponding to G has infinite degree, so is a non-algebraic Riemann surface. In fact, I have never really thought about what it looks like. It's fundamental group is the commutator subgroup of the fundamental group of C, which I believe is a free group of infinite rank. I don't think the field of meromorphic functions on this guy is what you want.</p> <p>2) On the other hand, the Hilbert class group $G$ of $K$ can be viewed as the Picard group of $\mathbb{Z}_K$, which classifies line bundles on $\mathbb{Z}_K$. This generalizes nicely: the Picard group of $C$ is an exension of $\mathbb{Z}$ by a $g$-dimensional complex torus $J(C)$, which has exactly the same abelian fundamental group as $C$ does: indeed their first homology groups are canonically isomorphic. $J(C)$ is called the <b>Jacobian</b> of $C$. </p> <p>3) It is known that every finite unramified abelian covering of $C$ arises by pulling back an isogeny from $J(C)$. </p> <p>So there are reasonable claims for calling either $G \cong \mathbb{Z}^{2g}$ and $J(C)$ the Hilbert class group of $C$. These two groups are -- canonically, though I didn't explain why -- Pontrjagin dual to each other, whereas a finite abelian group is (non-canonically) self-Pontrjagin dual. [This suggests I may have done something slightly wrong above.]</p> <p>As to what the Hilbert class field should be, the analogy doesn't seem so precise. Proceeding most literally you might take the direct limit of the function fields of all of the unramified abelian extensions of $C$, but that doesn't look like such a nice field. </p> <p>Finally, let me note that things work out much more closely if you replace $\mathbb{C}$ with a finite field $\mathbb{F}_q$. Then the Hilbert class field of the function field of that curve is a finite abelian extension field whose Galois group is isomorphic to $J(C)(\mathbb{F}_q)$, the (finite!) group of $\mathbb{F}_q$-rational points on the Jacobian.</p> http://mathoverflow.net/questions/10947/whats-the-analogue-of-the-hilbert-class-field-in-the-following-analogy/10955#10955 Answer by Felipe Voloch for What's the analogue of the Hilbert class field in the following analogy? Felipe Voloch 2010-01-06T20:12:07Z 2010-01-06T20:12:07Z <p>The absolute Galois group is $\pi_1( Spec K)$ and the Galois group of the maximal unramified extension of $K$ is $\pi_1( Spec R)$, where $R$ is the ring of integers of $K$ and the class group of $K$ is the abelianization of $\pi_1( Spec R)$ and the Hilbert class field is the field corresponding to this quotient.</p> <p>A curve has a function field and therefore an absolute Galois group, which is much bigger than the fundamental group and describes all branched covers of the curve. The fundamental group corresponds of course to unramified covers. Working over C gives you infinite groups even for the abelianization. If you work over a finite field, then the analogy is more precise.</p>