7
$\begingroup$

Historically (as I gather from Learning Class Field Theory: Local or Global First?), global class field theory was proved first, and then used to deduce local class field theory. But nowadays most treatments do the local theory first.

Can someone give me a summary of how the global-to-local argument goes, or a reference to where the argument appears?

I'm interested in seeing this as a toy model for how to think about the current state of Langlands, where it's still the case that many local statements (e.g. local Langlands for $GL_n$) are proved via global methods.

I think these proofs start off by globalizing the local situation, getting a corresponding global object on the other side, and then extracting a local component. The issue is then showing that this construction doesn't depend on the globalization. Can you prove local class field theory along these lines?

$\endgroup$
1
  • 3
    $\begingroup$ Probably not the answer you are looking for but, if you use the $t^n$-torsion in the Carlitz module to get abelian extensions of $\mathbb{F}_q(t)$ ramified only at $t$, you get the Lubin-Tate formal group for $\mathbb{F}_q((t))$. $\endgroup$ Dec 1, 2019 at 22:34

1 Answer 1

6
$\begingroup$

That's the approach taken in Lang's Algebraic Number Theory Springer GTM 110. Lang develops global class field theory, and then in Chapter XI Section 4 he finishes "the proof of the complete splitting theorem and derives local class field theory, describing the effect of the Artin map on the local component $k_v^*$ for a fixed $v$."

$\endgroup$
2
  • $\begingroup$ Thanks! This certainly answers the question, though the proof is more "direct" than what I had in mind in my last paragraph. $\endgroup$
    – user125639
    Dec 2, 2019 at 19:11
  • $\begingroup$ @user125639 Sorry, I didn't mean this as an answer to the question in your last paragraph. I have no idea if or how this approach can be applied to the GL$_n$ case for $n\ge2$. I was just answering your initial request for a reference where the GL$_1$ argument appears. $\endgroup$ Dec 2, 2019 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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