In the book's Algebraic Number theory, Ian StewarT, Third edition (page 51-52), has the following propositions:
Theorem 2.20: Let $G$ be an additive subgroup of $\mathfrak{O}_K$ of rank equal to the degree of $K$, with $\mathbb{Z}$-basis $\{\alpha_1,\ldots,\alpha_n\}$. Then $|\mathfrak{O}_K/G|^2$ divides $\Delta[\alpha_1,\ldots,\alpha_n]$.
Proposition 2.21: Suppose that $G\neq{\mathfrak{O}_K}$. Then there exists an algebraic integer of the form
$$\frac{1}{p}(\lambda_1\alpha_1+\cdots+\lambda_n\alpha_n) \ (* )$$ where $0\leq{i}\leq{p-1}$, $\lambda_i\in\mathbb{Z}$, and $p$ is a prime such that $p^2$ divides $\Delta_G$.
Obs:
1) $K$ is a number field, $\mathfrak{O}_K$ is of ring of algebraic integers.
2) If $\sigma_i:K\to{\mathbb{C}}$, with $i=1,\ldots,n$ ($n=[K:\mathbb{Q}]$) are monomorphisms, then $\Delta_G=\Delta[\alpha_1,\ldots,\alpha_n]=det(\sigma_i(\alpha_j))^2$.
We may use proposition 2.21 as the basis of a trial-and-error search for algebraic integers in $\mathfrak{O}_K$ but not in $G$, because there are only finitely many posibilities. The idea is:
a) Start with a initial guess $G$ for $\mathfrak{O}_K$
b) Compute $\Delta_G$.
c) For each prime $p$ whose square divides $\Delta_G$, test all numbers of the form $(*)$ to see which are algebraic integers.
d) If any new integers arise,enlarge $G$ to a new $G´$ by adding in the new number (and divide $\Delta_G$ by $p^2$ to get $\Delta_G´$)
My question is: In d), why $\displaystyle \Delta_G´=\frac{ \Delta_G }{p^2}$?
I hope I can clarify this doubt.
Thanks you all.
If there is an element in $\frak O$ of the form $(*)$ then there is an adapted $\mathbf Z$-basis for $G'$ and its subgroup $G$, i.e., there is a $\mathbf Z$-basis $\beta_1,\ldots,\beta_n$ of $G'$ such that $p\beta_1,\beta_2,\ldots,\beta_n$ is a $\mathbf Z$-basis of $G$. We may assume that $$\alpha_1=p\beta_1,\alpha_2=\beta_2,\ldots,\alpha_n=\beta_n.$$ Then $$ \Delta_{G}=\det(\sigma_i(\alpha_j))^2=(p\det(\sigma_i(\beta_j)))^2=p^2\det(\sigma_i(\beta_j))^2=p^2\Delta_{G'} $$ since $\sigma_i(\alpha_1)=\sigma_i(p\beta_1)=p\sigma_i(\beta_1)$ for all $i$. Hence $$ \Delta_{G'}=\frac{\Delta_G}{p^2}. $$