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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.

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

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