Reducing Congruences modulo an ideal of a ring of algebraic integers to congruences over $\mathbb{Z}$ Let $\mathbb{F}$ be a number field of degree $d$, let $\mathcal{O}_{\mathbb{F}}$ be the ring of $\mathbb{F}$-integers, and let $\mathfrak{I}$ be an ideal of $\mathcal{O}_{\mathbb{F}}$ for which the quotient ring $\mathcal{O}_{\mathbb{F}}/\mathfrak{I}$ is finite—say, containing $\varrho$ distinct equivalence classes, which I'll label as $\mathfrak{I}_{0},\ldots,\mathfrak{I}_{\varrho-1}$, where $\mathfrak{I}_{0}$ is the equivalence class of $\mathfrak{I}$ itself. Next, let $\gamma_{1},\ldots,\gamma_{d}\in\mathcal{O}_{\mathbb{F}}$ be a basis of $\mathcal{O}_{\mathbb{F}}$ over $\mathbb{Z}$, and write elements of $\mathcal{O}_{\mathbb{F}}$ as $d$-tuples $\left(n_{1},\ldots,n_{d}\right)\in\mathbb{Z}^{d}$ with respect to this basis.
Now, suppose: $$x=\sum_{\ell=1}^{d}n_{\ell}\gamma_{\ell}\in\mathcal{O}_{\mathbb{F}}$$ satisfies $x\overset{\mathfrak{I}}{\equiv}\mathfrak{I}_{j}$ for some $j\in\left\{ 0,\ldots,\varrho-1\right\}$  (i.e., there is an element $x_{j}\in\mathfrak{I}_{j}$ so that $x-x_{j}\in\mathfrak{I}$). My question is: is there a general procedure for converting the congruence $x\overset{\mathfrak{I}}{\equiv}\mathfrak{I}_{j}$ to a congruence (or system thereof) involving the $n_{\ell}$s? Also, is there a choice of basis $\gamma_{1},\ldots,\gamma_{d}$ which minimizes the amount of work needed to do this conversion?
For example, let $\mathbb{F}=\mathbb{Q}\left(\sqrt{2}\right)$, $\mathcal{O}_{\mathbb{F}}=\mathbb{Z}\left[\sqrt{2}\right]$, and let $\mathfrak{I}=\left\langle \sqrt{2}\right\rangle$. Choose $\gamma_{1}=1$ and $\gamma_{2}=\sqrt{2}$, and let $x=a+b\sqrt{2}$ be arbitrary. Since $\left\langle 2\right\rangle$  is a subring of $\left\langle \sqrt{2}\right\rangle$, it follows that everything congruent to $0$ mod $2$ will also be congruent to $0$ mod $\sqrt{2}$. Now, since $2\in\mathfrak{I}$, it follows that $a+b\sqrt{2}\overset{\sqrt{2}}{\equiv}a$. If $a$ is even, then $a\overset{\sqrt{2}}{\equiv}0$. If $a$ is odd, then $a\overset{2}{\equiv}1$. Since $\left\langle \sqrt{2}\right\rangle$  contains the subring $\left\langle 2\right\rangle$, the congruence $a\overset{2}{\equiv}1$ then “lifts” to the congruence $a\overset{\sqrt{2}}{\equiv}1$.
Consequently, $a+b\sqrt{2}$ is congruent to $0$ mod $\sqrt{2}$ if and only if $a$ is congruent to $0$ mod $2$, and $a+b\sqrt{2}$ is congruent to $1$ mod $\sqrt{2}$ if and only if $a$ is congruent to $1$ mod $2$. Since this exhausts all possible equivalence classes of $a$ mod $2$, this then completely classifies the ways in which $a+b\sqrt{2}$ can fall into different equivalence classes mod $\sqrt{2}$.
I can do this simple case, but anything of dimension three or higher just boggles my mind. Algebra is not my strong suit by any stretch of the imagination, so any help in dealing with the general problem would be much appreciated!
 A: Here is an idea that reduces the problem to solving a system of linear Diophantine equations. I would not be surprised if there are better methods, I'd be interested to see.
The ring $\mathcal{O}_\mathbb{F}$ is Noetherian, so every ideal is finitely-generated. Let $\mathfrak{I} \subset \mathcal{O}_\mathbb{F}$ be such an ideal, and let the generators be $\alpha_1
, \ldots, \alpha_n$ (actually, since $\mathcal{O}_\mathbb{F}$ is also a Dedekind domain, every ideal can be generated by at most $2$ elements, but we don't need this fact here).
An element $x \in \mathcal{O}_\mathbb{F}$ is contained in $\mathfrak{I}$ iff there exist $\beta_i \in \mathcal{O}_\mathbb{F}$ such that $x = \sum\limits_{i=1}^n \beta_i\alpha_i$. Write each $\beta_i$ as a linear combination of the integral basis of $\mathcal{O}_\mathbb{F}$: $$\beta_i = \sum\limits_{j=1}^d b_{ij} \gamma_j\text{ , where }b_{ij} \in \mathbb{Z}$$
Also write each $\alpha_i\gamma_j$ as a linear combination of the $\gamma_k$: $$\alpha_i\gamma_j = \sum\limits_{k=1}^da_{ijk}\gamma_k\text{ , where }a_{ijk} \in \mathbb{Z}$$
Hence we have $$x = \sum\limits_{i=1}^n\sum\limits_{j=1}^db_{ij}\alpha_i\gamma_j = \sum\limits_{i=1}^n\sum\limits_{j=1}^d\sum\limits_{k=1}^db_{ij}a_{ijk}\gamma_k = \sum\limits_{k=1}^d\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^da_{ijk}b_{ij}\right)\gamma_k$$
Note that the coefficients $a_{ijk} \in \mathbb{Z}$ are known in principle, given the generators $\alpha_1, \ldots, \alpha_n$ of $\mathfrak{I}$, and the integral basis $\gamma_1, \ldots, \gamma_d$ of $\mathcal{O}_\mathbb{F}$.
So if we are given $x = \sum\limits_{k=1}^dc_k\gamma_k$, we can determine whether $x \in \mathfrak{I}$ by checking if the system of linear equations $$\sum\limits_{i=1}^n\sum\limits_{j=1}^da_{ijk}b_{ij} = c_k\text{ , where }k = 1, \ldots, d$$ has solutions $b_{ij}$ over the integers. This ends up being a system of $d$ linear Diophantine equations in $nd$ unknowns, which can be solved with the Smith normal form.

We can apply this to your example, where $\mathbb{F} = \mathbb{Q}(\sqrt{2})$ and $\mathcal{O}_\mathbb{F} = \mathbb{Z}[\sqrt{2}]$. Let's look at the ideal $\mathfrak{I} = \langle \sqrt{2} \rangle$. An integral basis for $\mathcal{O}_\mathbb{F}$ is given by $\{1, \sqrt{2}\}$. Then we have $\alpha_1 = \sqrt{2}$, $\gamma_1 = 1$ and $\gamma_2 = \sqrt{2}$. We find $\alpha_1\gamma_1 = \sqrt{2}$ and $\alpha_1\gamma_2 = 2$. Hence $a_{111} = 0$, $a_{112} = 1$, $a_{121} = 2$ and $a_{122} = 0$.
Given $x = c_1 + c_2\sqrt{2}$, membership in $\mathfrak{I}$ is determined by solving the system:
$$2b_{12} = c_1$$
$$b_{11} = c_2$$
This has integer solutions for $b_{ij}$ iff $c_1$ is even. Therefore $c_1 + c_2\sqrt{2} \in \langle\sqrt{2}\rangle$ iff $c_1$ is even, which is what you already concluded too.

So far I only talked about how to determine membership in the ideal $\mathfrak{I}$, but not membership in its cosets. But using the procedure as above, we end up with a system of linear Diophantine equations. If this system has no solutions, then we know that $x$ lies in another coset of $\mathfrak{I}$. But I think we can read it off from the Smith normal form and the congruence class of the $c_k$ modulo the diagonal entries of this matrix. I'll come back to add details on this later if I can think of it.
