That's it. I want to do this quotient, $\mathbb Z [\sqrt{-11}] / (1+\sqrt{-11})$. My first idea was to see which elements are in the ideal:
$(a+b\sqrt{-11})(1+\sqrt{-11}) = a+ a\sqrt{-11} + b\sqrt{-11} -11b$
So I want to make and application who send this elements to $0$, and then see the Kernel of the application and apply the Isomorphism Theorem.
Any ideas?
One could try to describe the ideal completely. It will end up like $$ a + b\sqrt{-11}\in (1+\sqrt{-11})\iff \exists x, y\in \Bbb Z( a = x-11y\land b = x+y) $$ In other words, $(1+\sqrt{-11})$ contains all elements of the form $(x-11y) + (x+y)\sqrt{-11}$ for integers $x, y$.
There are two main approaches here. One is to try to find a nice way of uniquely representing the elements of the quotient in a way that makes it easier to see what ring we're dealing with. The other is to do some third isomorphism theorem trickery to end up with a much easier calculation.
Finding representatives
We know that $1 + \sqrt{-11}$ is in the ideal, so any congruency class has a representative which is a pure integer. In other words, $$ a + b\sqrt{-11} + (1 + \sqrt{-11}) = a-b + (1+\sqrt{-11}) $$ Now note that $12$ is in our ideal (it is $(1 + \sqrt{-11})(1 - \sqrt{-11})$). So we can subtract multiples of $12$ from these representatives without changing what congruency class they represent. So the set $$ \{a + (1 + \sqrt{-11}) \mid 0\leq a<12\} $$ of representatives covers all of the quotient ring. Are we done?
Consider the map $\varphi: \Bbb Z[\sqrt{-11}] \to \Bbb Z/(12)$ given by $$ \varphi(a + b\sqrt{-11}) = a-b + (12) $$ We have discovered that the kernel of this map contains $(1 + \sqrt{-11})$ (it is also not difficult to confirm through pure calculation: $\varphi(1 + \sqrt{-11}) = 0 + (12)$). But is that all of the kernel?
Take an element in the kernel of $\varphi$, which is to say an element $a + b\sqrt{-11}\in \Bbb Z[\sqrt{-11}]$ such that $a - b$ is divisible by $12$. Then consider $$ x = \frac{a + 11b}{12} = \frac{a-b}{12} + b\\ y = \frac{b-a}{12} $$ We see that $a - b$ being divisible by $12$ implies that $x$ and $y$ are integers, and we also see that $a = x -11y$ and $b = x+y$. Thus, by the general form of an element of $(1+\sqrt{-11})$ seen above, we see that $a + b\sqrt{-11}$ is indeed contained in $(1 + \sqrt{-11})$, and we are done.
Isomorphism theorem
Write $\Bbb Z[\sqrt{-11}]$ itself as a quotient ring: $\Bbb Z[x]/(x^2+11)$ (where $x$ takes the role of $\sqrt{-11}$). The ring we are ultimately after is isomorphic to $\Bbb Z[x]/(x^2+11, x+1)$.
It turns out that we can divide by these two generators one at a time in either order (as a consequence of the third isomorphism theorem). The problem statement implies doing it in one order (first $x^2+11$, then $x+1$), but there is no reason to stick to that.
This makes things a lot easier: $\Bbb Z[x]/(x+1)$ is just (isomorphic to) $\Bbb Z$, and the quotient map sends $x$ to $-1$. Now we can see what happens to the other generator: it gets sent to $(-1)^2+11 = 12$. So the final ring is isomorphic to $\Bbb Z/(12)$.
Alternately, it is not difficult to show that $12\in (x^2+11, x+1)$ directly: $12 = x^2+11 - (x-1)(x+1)$. Then note that $x^2 + 11\in (12, x+1)$ by basically the same calculation, proving that $(x^2+11, x+1) = (12, x+1)$. These two generators are much easier to divide out by, and we still see that the final result ends up being $\Bbb Z/(12)$.
You can write $(a+b\sqrt{-11})(1+\sqrt{-11}) = a+ a\sqrt{-11} + b\sqrt{-11} -11b$ as $$ \begin{pmatrix} a' \\ b' \end{pmatrix} = \begin{pmatrix} 1 & -11 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} $$ The matrix has determinant $12$ and so the quotient ring has (probably) order $12$.
Since $a - 11b \equiv a + b \bmod 12$, to kill them both try $$ \mathbb Z [\sqrt{-11}] \to \mathbb Z_{12}, \qquad a+b\sqrt{-11} \mapsto a-b \bmod 12 $$ This map is clearly surjective and additive. Check that it is multiplicative. Then prove that the kernel is $\langle 1+\sqrt{-11} \rangle$.
Write $\,R = \Bbb Z[w],\,\ \bar R = R/(1\!+\!w),\ w = \sqrt{-11},\,$ so $\,N(w) = w\bar w = \color{#c00}{12}$.
$h: \Bbb Z \to \bar R\,\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}\,$ by $\!\bmod\, 1\!+\!w\!:\ \, w\equiv -1\,\Rightarrow\, a\!+\!bw\equiv a\!-\!b\in\Bbb Z$
$\color{#c00}{I := \ker h = 12\,\Bbb Z}\ $ follows immediately by means of $\,\rm\color{#90F}{rationalizing}\,$ a denominator
$ n\in I\!\!\iff\!\! 1\!+\!w\mid n\ \, {\rm in}\, R\!$ $\iff\!\! \dfrac{n}{1\!+\!w}\in R\!$ $\color{#90f}{\overset{\large \rm\ rat}\iff}\! \dfrac{n(1\!-\!w)}{\color{#c00}{12}}\!\in\! R\!$ $\iff\! \color{#c00}{12\mid n}\ \,{\rm in}\,\ \Bbb Z$
Thus $\, \color{#0a0}{\bar R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{12\,\Bbb Z}\ $ by the First Isomorphism Theorem.
Remark $ $ We tested divisibility by $\color{#90f}{\textit{rationalizing}}$ the denominator in order to reduce division by an algebraic irrational $\,1+w\,$ to a simpler division by an integer (its norm $=12)$. This is a special case of the method of simpler multiples. Same proof works for $\,w=a+bi$ when $\gcd(a,b)=1$.
Another general view you might find illuminating arises from rewriting the ideal as a module in Hermite normal form: $\ I = (1\!+\!w) = (12,1\!+\!w) = 12\Bbb Z + (1\!+\!w)\Bbb Z.\,$ But it is trivial to test module membership given such a triangularized basis, namely $$\begin{align} a\!+\!bw = a\!-\!b +b(1\!+\!w)&\in I = 12\Bbb Z + (1\!+\!w)\Bbb Z\\ \iff\ a\!-\!b&\in I\\ \iff\ a\!-\!b &\in 12\Bbb Z \iff 12\mid a\!-\!b \end{align}\qquad$$
Further this shows that $\, a\!+\!bw\bmod I\, =\, a\!-\!b\,\bmod 12.\ $
The criterion generalizes to an ideal test for modules $\rm\,[a,b\!+\!c\:\!\omega]\,$ in the ring of integers of a quadratic number field, e.g. see section 2.3 Franz Lemmermeyer's notes linked here..
This is a special case of module normal forms that generalize to higher degree number fields, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's $ $ A Course in Computational Number Theory.
