I am trying to understand a proof of the following statement, but I do not understand it completely. Could you please tell me, if my thinking is correct.
If K is an algebraic number field and $\Lambda \subset \Lambda '$ are two non-zero finitely generated $\mathcal{O}_{K}$-submodules of K, then we have for the discriminant $d(\Lambda)=(\Lambda ' : \Lambda)^2 * d(\Lambda ')$
I know that $\Lambda$ and $\Lambda '$ both admits a $\mathbb{Z}$-basis, so I can write every element of $\Lambda$ as a linear combination of some basis of $\Lambda '$ and $\exists A \in M_{n}(\mathbb{Z}) : \Lambda=A \Lambda '$.
In the next step of the prove we take a look at $\Lambda '/ \Lambda \cong \mathbb{Z}^n / A \mathbb{Z}^n$. Is this possible because we can use the theorem of finitely generated modules over principal domains, and $\Lambda$ and $\Lambda '$ are free $\mathbb{Z}$-modules?
How exactly looks an element in $\mathbb{Z}^n / A \mathbb{Z}^n$? The proof states that $\mathbb{Z}^n / A \mathbb{Z}^n$ is a finitely generated abelian group. I do not think this is difficult to see, but I do not have any idea what exactly $\mathbb{Z}^n / A \mathbb{Z}^n$ is. Is the multiplication of $A$ and $\mathbb{Z}^n$ defined component-wise?
Now by the theorem of finitely generated abelian groups we get $\Lambda '/ \Lambda \cong \mathbb{Z} / d_{1} \mathbb{Z} \oplus ...\oplus \mathbb{Z} / d_{n} \mathbb{Z} \oplus \mathbb{Z}^r$ Why exactly do we know that $r=0$?
Next we can use the theorem of the Smith normal form, because $A \in M_{n}(\mathbb{Z})$, and find $B, B'$ such that $BAB' = \begin{pmatrix} d_1 & 0 & ... & 0 \\ 0 & d_2 & ... & 0 \\ ... & ... & ... & ... \\ 0 & 0 & ... & d_n \end{pmatrix}$ with $d_{1}|d_{2}|...|d{n}$. Now we have $det(A) = d_{1}d_{2}...d_{n} = | \mathbb{Z}^n /A \mathbb{Z}^n| = ( \Lambda ' : \Lambda)$ This is clear for me.
So last but not least it follows that $d( \Lambda) = d(A \Lambda ') = det(A)^2d( \Lambda ') = ( \Lambda ' : \Lambda )^2d( \Lambda ')$ and here I do not get, why we have the 2nd $=$.
Thank you very much!