Learning class field theory I found this theorem, but I can't prove it or find the solution. I'll be glad to any help.

Let $L$ and $M$ be abelian extensions of $K$. $L \subset M$ if and only if there is a modulus $\mathfrak m$, divisible by all primes of $K$ ramified in either $L$ or $M$, such that $$ P_{K,1}(\mathfrak m)\subset \ker(\Phi_{M/K,\mathfrak m}) \subset \ker(\Phi_{L/K,\mathfrak m}).$$

$\Phi_{M/K,\mathfrak m}$ - is Artin map for modulus $\mathfrak m$. $P_{K,1}$ is a subgroup of the group of fractions ideals, generated by principal $\alpha \mathcal O_K$-ideals, where $\alpha$ satisfies $\alpha \equiv 1 \pmod{\mathfrak m_0}$ and $\sigma (\alpha) > 0$ for every real infinite prime $\sigma$ dividing infinite part of $\mathfrak m.$

It's quite easy to prove that $L \subset M$ implies $$ P_{K,1}(\mathfrak m)\subset \ker(\Phi_{M/K,\mathfrak m}) \subset \ker(\Phi_{L/K,\mathfrak m}).$$ But I don't know how to prove another implication.

  • 1
    $\begingroup$ The compositum of class fields corresponds to the intersection of the associated ideal groups. $\endgroup$ – franz lemmermeyer May 12 '13 at 11:35

The way I know how to answer this question involves works with ideles rather than ideals. As long as $P_{1,m}$ is contained in the kernel of the Artin map $I(m) \rightarrow Gal(L/K)$, the global norm index inequality tells you that this kernel is exactly $P_{1,m} \mathfrak N_{L/K}(m)$, where $\mathfrak N_{L/K}(m)$ is the group of norms from $L$ which are relatively prime to $m$.

From an approximation argument (for example found in Lang) what you get is a well defined Artin map on $\mathbb{I}_K$ (the ideles of $K$) to $Gal(L/K)$ whose kernel is $H_L := K^{\ast}N_{L/K}(\mathbb{I}_L)$. Similarly you get an Artin map on $\mathbb{I}_K \rightarrow Gal(M/K)$ whose kernel is $H_M := K^{\ast} N_{L/K}(\mathbb{I}_M)$. The containment of kernels of ideals translates to a containment of kernels of ideles: $$H_M \subseteq H_L$$ Without assuming any containment of $H_M, H_L$, you can check that the kernel of the Artin map on the composite field $LM$ is the $H_M \cap H_L$. Since in this case $H_M \cap H_L = H_M$, we have by the global norm index equality that $$[ML : K] = [\mathbb{I}_K : H_M \cap H_L] = [\mathbb{I}_K : H_M] = [M : K]$$ so $ML = M$, or $L \subseteq M$. I left out a lot of details.

You may be able to work out the corresponding argument just within the ideals, but ideles are really a better way to organize the abelian extensions.


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.