let $K = \mathbb{Q}(\sqrt{-5})$ and $L = K(i)$. $\mathcal{O}_K$ is the ring of integers of K.
I would like to show that the kernel of the Artin-Map $\phi_{L/K}: I_K \rightarrow Gal(L/K)$ is $P_K$, where $I_K$ denotes the group of all fractional Ideals from the ring of integers of $K$ and $P_K$ denots the group of all prinicipal ideals.
Because I´m new to this topic, I want to do it without class field theory, but recently I read a lot about the frobenius-element/artin-symbol of a prime ideal.
So far I thought about the following: I guess I can compute the frobenius-element/artin-symbol of every prime $\mathfrak{P}$ from $\mathbb{Q}(\zeta_{20})$ which lies about a integer prime $p \neq 2,5$. I think this can be useful since $K(i) \subset \mathbb{Q}(\zeta_{20})$. I also know that $2\mathcal{O}_K = (2, 1 +\sqrt{-5})^{2}$ and $5\mathcal{O}_K = (\sqrt(-5))^2$.
Since $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is abelian, the Artin-Symbol of $\mathfrak{P}$ depends only on the underlying prime integer p, for which $(p) = \mathfrak{P} \cap \mathbb{Z}$.
The Artin-symbol of every prime integer $p \equiv 1,3,7,9,11,13,17,19\ mod\ 20$ should be the field-automorphism $\sigma_p$ which maps $\zeta_n$ to $\zeta_n^{p}$.
Further, $<\sigma_3>$ fixes $K$ and $<\sigma_9>$ fixes $L$. Now I know
$p$ splits completely in $K$ resp. $L$ iff $\sigma_p$ fixes $K$ resp. $L$.
Now let $p \equiv 11,13,17,19\ mod\ 20$ a prime integer.
Then $p$ doesn´t split in $K$, which mean (in this case) there is a prime ideal $\mathfrak{p}$ from $\mathcal{O}_K$ lying over $p$ with inertial degree $f(\mathfrak{p}|p) = 2$.
Since $p$ doesn´t split completely in $L$ either, it can only split into one or two prime ideals. But $|<\sigma_{p}|_{L}>| = 2$, where $\sigma_{p}|_{L}$ is the restriciton to $L$. That means that $p$ splits into 2 prime ideals in $L$. That means there a prime ideals $\mathfrak{P}_1,\mathfrak{P}_2$ in $L$ with $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2$. But also $p\mathcal{O}_K = \mathfrak{p}$ with $\mathfrak{p}$ is a prime ideal in $K$. Then I can see that $f(\mathfrak{P}_i|\mathfrak{p}) = 1$, which means that $\mathfrak{p}$ splits completely in $L$. Is this right so far?
Now let $p \equiv 1,3,7,9\ mod\ 20$ be a prime integer.
$p$ splits completely in $K$, which means there a prime ideal $\mathfrak{p}_1,\mathfrak{p}_2$ in $K$ with $p\mathcal{O}_K = \mathfrak{p}_1\mathfrak{p}_2$. So $f(\mathfrak{p}_i|p) = 1$.
Now for $p \equiv 1,9\ mod\ 20$.
$p$ splits completely in $L$, which means there a prime ideal $\mathfrak{P}_1,\mathfrak{P}_2,\mathfrak{P}_3,\mathfrak{P}_4$ in $L$ with $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2\mathfrak{P}_3\mathfrak{P}_4$. So $f(\mathfrak{P}_i|p) = 1$ and hence $f(\mathfrak{P}_i|\mathfrak{p}_j) = 1$.
So every prime ideal $\mathfrak{p}$ in $K$, lying over a prime $p \equiv 1,9\ mod\ 20$ splits completely in $L$, is that right?
When I do the same for $p \equiv 3,7\ mod\ 20$ then I get that every prime ideal $\mathfrak{p}$ in $K$, lying over such a prime integer $p$, doesn´t split in $L$, because $f(\mathfrak{p}|p) = 1$, but also $|<\sigma_{p}|_{L}>| = 2$, which means $p\mathcal{O}_L = \mathfrak{P}_1\mathfrak{P}_2$. Therefore $f(\mathfrak{P}|\mathfrak{p}) = 2$.
According to my calculations only the prim ideals $\mathfrak{p}$ in $K$, which lie over a prime integer $p \equiv 3,7\ mod\ p$ doesn´t split in $L$. can somebody confirm this?
Now, except for those prime ideals lying over 2 and 5, I can tell which prime ideals (dependent on the underlying prime integer) in $K$ splits completely in $L$, which means mapping to the 1 in $Gal(L/K).
Now, how can I check that those prime ideals are exactly the principal prime fractional ideals? And which prime ideals generate $P_K$?
Thanks in advance Julian
