Finding the kernel of a morphism : Galois Theory and Decomposition Field I am currently doing an internship in a research laboratory ( I am in my third year of Bachelor ) and I'm really struggling with the things I have to do.
For instance, here's something I'm having trouble with.
Let $L$ be a finite Galois extension of $\mathbb{Q}$, $O_L$ its ring of integers, and $I$ an ideal of $O_L$. Let $K$ be the decomposition field of $I$. Let $R=I\cap O_K$. Suppose $n=[L : \mathbb{Q} ]$ and $g=[ K : \mathbb{Q} ]$. Suppose also that we have a basis $(b_i)_{1 \leq i \leq n/g}$ of $O_L$ over $O_K$.
In order to find an isomorphism between $(O_K/R)^{n/g}$ and $O_L/I$, I wanted to proceed like that :
$(O_K)^{\frac{n}{g}} \overset{f_1}{\longrightarrow} O_L \overset{f_2}{\longrightarrow} O_L /I$
$f_1$ being : $ (x_1,\cdots , x_{n/g}) \longrightarrow \sum\limits_{i=1}^{n/g} x_i b_i$
and $f_2$ being the canonical surjection.
In order to find my isomorphism, I need to prove that the kernel of the composition of $f_1$ and $f_2$ is $R^{\frac{n}{g}}$.
The Kernel of $f_2$ being $I$, what's left to prove is that :
$R^{\frac{n}{g}}=\{ x\in (O_K)^{\frac{n}{g}} ~:~ f_1(x)\in I \}$
The $\subset$ of the equality is easy, but I can't prove the $\supset$.
Please forgive my mistakes, I'm still learning, and English isn't my first language.
Thank you for your help !
 A: If $K$ is the subfield of $L$ fixed by $\{ \sigma\in Gal(L/\Bbb{Q}), \sigma(I)\subset I\}$
then try with $O_L=\Bbb{Z}[i], I=(1+i)^3,O_K=\Bbb{Z},I\cap O_K=(4)$,
As a group $O_L/I\cong C_4\times C_2$
($C_4$ is the subgroup generated by $1$ and $C_2$ the subgroup generated by $1+i$)
If $I=P^d$ and $P$ is unramified then yes your statement holds, this follows from $|O_L/I|=|O_K/(P^d\cap O_K)|^{f(P)}$ where $n=efg,e=1$ so the kernel of your surjective map $O_K^{n/g}\to O_L/I$ can't be larger than $(O_K\cap I)^{n/g}$.
When $I$ is a product of distinct unramified primes powers it should often fail, say when the decomposition group is not the same for each prime.
A: Question: "Please forgive my mistakes, I'm still learning, and English isn't my first language. Thank you for your help!"
Answer: It seems you may reduce to the case of $I$ being a power of a prime ideal. Since $I \subseteq A:=\mathcal{O}_L$ it follows there are maximal ideals
$\mathfrak{p}_i$ for $i=1,..,n$ with
$$I=\prod_i \mathfrak{p}_i^{l_i}.$$
You get a diagram of exact sequences $\require{AMScd}$
\begin{CD}
 \mathcal{O}_L @>>> \mathcal{O}_L/I @>>>  \oplus_i\mathcal{O}_L/\mathfrak{p}_i^{l_i} \\
    @V V V @VV V  @VVV \\
    \mathcal{O}_L @>>> \mathcal{O}_L/\mathfrak{p}_i^{l_i} @>>> \mathcal{O}_L/\mathfrak{p}_i^{l_i}
\end{CD}
where the middle vertical arrow is the canonical map and the rightmost vertical arrow is the projection map. In the Neukirch book on ramification theory they develop the theory of decomposition fields and decomposition groups in the case when $I:=\mathfrak{p_1}$ and $l_1=1$. A good idea could be to try to develop this theory for powers of prime ideals.
