Let $K\subseteq L$ be a two number fields with ring of integers $\mathcal O_K$ and $\mathcal O_L$ respectively. And let $I(K)$ and $I(L)$ (resp. $P(K)$ and $P(L)$) the fractional ideals (resp. the principal fractional ideals) of $K$ and $L$ respectively.
- My first question is if the following map is well-defined
$$I(K)\to I(L),\ \mathfrak a\mapsto \mathfrak a\mathcal O_L.$$
My progress is as follows.
First, if $\mathfrak a=\alpha\mathfrak b$ is a fractional ideal of $K$ with $\alpha \in K^\times$ and $\mathfrak b$ is a ideal of $\mathcal O_K$, then $\mathfrak a \mathcal O_L$ is fractional ideal because $\mathfrak a \mathcal O_L=\alpha \mathfrak b\mathcal O_L$ and $\mathfrak b\mathcal O_L$ is a ideal of $\mathcal O_L$. Is correct?
Now, this gives that $P(K)$ maps to $P(L)$, and thus we have a well-defined map between the class groups of $K$ and $L$
$$\varphi: C(K)\to C(L).$$
- My second question is if the map $\varphi$ is injective.
For this, first check the injectivity of the map $I(K)\to I(L)$.
Suppose that $\mathfrak p \mathcal O_L=\mathfrak q\mathcal O_L=\prod_i\mathfrak B_i$, with $\mathfrak p$ and $\mathfrak q$ prime ideals of $\mathcal O_K$ and the $\mathfrak B_i's$ are prime ideals of $\mathcal O_L$.
The intersection of the prime ideals $\mathfrak B_i's$ with $\mathcal O_K$ is a prime ideal that contains $\mathfrak p$ and $\mathfrak q$, and thus $\mathfrak p=\mathfrak q$. This proves the injectivity for prime ideals, therefore the injectivity for all, because prime ideals generate the fractional ideals.
This correct my two arguments?
Thanks you all
The map you have defined is called the transfer of ideal classes. It is well defined, and its kernel is called the capitulation kernel. It is easy to see that if an ideal class of order $k$ capitulates, then $k$ must divide the degree $(L:K)$ of the extension. But in general the transfer map is not injective; in particular the transfer map is the trivial map when $L$ is the Hilbert class field of $K$ by Furtwängler's principal ideal theorem.