Dtyjlui

This might be a very basic question for some of you. Indeed in $textbf Z$, it's very easy. For example, $textbf Z / langle 2 rangle$ consists of $langle 2 rangle$ and $langle 2 rangle + 1$. Obviously just two elements. In general, if $p$ is a positive prime in $textbf Z$, then $textbf Z / langle p rangle$ consists of one principal ideal and $p - 1$ cosets.

I guess it's also easy in imaginary quadratic integer rings, since we can visualize them in the complex plane, e.g., $textbf Z[i] / langle 1 + i rangle$ consists of $langle 1 + i rangle$, $langle 1 + i rangle + 1$ and $langle 1 + i rangle + i$... wait a minute, three elements? I'm not sure that's quite right.

And I really have no idea how to go about, say, $textbf Z[sqrt{14}] / langle 4 + sqrt{14} rangle$. To say nothing of something like $textbf Z[sqrt{10}] / langle 2, sqrt{10} rangle$.

Given a ring $R$ of algebraic integers of degree $2$, and a prime ideal $mathfrak P$, how do you determine how many elements there are in $R / mathfrak P$?

For principal ideals, it's easy: if $L$ is a number field with ring of integers $mathcal O_L$, and if $(a)$ is an ideal in $mathcal O_L$, then the number of elements in $mathcal O_L / (a)$ is $|N_{L / mathbb Q} (a)|$, where $N_{L / mathbb Q}(a)$ denotes the norm of the element $a$.

For a quadratic number field $mathbb Q(sqrt{d})$, the norm of an element $x + y sqrt{d}$ is simply $x^2 - dy^2$.

So the number of elements in $mathbb Z[sqrt{14}] / (4 + sqrt{14})$ is $| 4^2 - 14 times 1^2 | = 2$.

Similarly, the number of elements in $mathbb Z[i] / (1 + i)$ is $|1^2 + 1 times 1^2 | = 2$. (You double counted: $1 + (1 + i)$ and $i + (1 + i)$ are the same thing, since $1 - i in (1 + i)$ by virtue of the fact that $-i(1 + i) = 1 - i$.)

As for your final example, $mathbb Z[sqrt{10}] / (2, sqrt{10})$, it might be easier to use an integral basis.

It is well known that ${ 1, sqrt{10} }$ is an integral basis for $mathbb Z[sqrt{10}]$, which is to say that every element in $mathbb Z[sqrt{10}]$ can be written in the form $a + bsqrt{10}$ for a unique choice of $a, b in mathbb Z$. Therefore, the elements in $(2, sqrt{10})$ are precisely the elements of the form

$$ (u + vsqrt{10})times 2 +(w + zsqrt{10})times sqrt{10} = (2u + 10z) + (2v + w) sqrt{10}$$
where $u, v, w, z in mathbb Z$. And if you think about it, this includes precisely those elements $a + b sqrt{10}$ in $ mathbb Z[sqrt{10}]$ where $a$ is even. Hence $mathbb Z[sqrt{10}] / (2, sqrt{10})$ contains two cosets: $(2, sqrt{10})$ and $1 + (2, sqrt{10})$.

I hope these examples provide you with enough techniques to be getting on with....

$$Bbb Z[sqrt{14}]/(4+sqrt{14}) = Bbb Z[X]/(X^2-14,X+4) = Bbb Z[X]/(2,X+4) = Bbb Z[X]/(2,X) = Bbb F_2$$

$$Bbb Z[sqrt{10}]/(2,sqrt{10}) = Bbb Z[X]/(X^2-10, 2, X) = Bbb Z[X]/(2, X) = Bbb F_2$$

The trick is to move the situation to computing ideals in $Bbb Z[X]$.

I'm afraid there is no general approach if not going through the ring of integers of the number field. Recall that for a number field $K$ of degree $n$ over $mathbf Q$, the norm $N(x)$ of an element $xin K^*$ is defined as the product $s_1(x)...s_n(x)$, where the $s_i$ are the $n$ distinct embeddings of $K$ into $mathbf C$. If $x$ is a non zero element of the ring of integers $R$ of $K$, it is known that $mid N(x)mid $ is finite, equal to card $(R/Rx)$. This is easily shown by using the fact that $R$ is a free $mathbf Z$-module of rank $n$ and, because $mathbf Z$ is a PID, there exists a $mathbf Z$-basis $(e_1 ,..., e_n)$ of $R$ and elements $c_j$ of $mathbf Z$ s.t. $(c_1 e_1,..., c_n e_n)$ is a $mathbf Z$-basis of $Rx$. Thus, to get card $(R/Rx)$, it suffices to compute $N(x)$, which is, up to a sign, the constant term of the minimal polynomial of $x$ over $mathbf Q$. Warning : if $K=mathbf Q (x)$, then $mathbf Z [x]$ is contained in, but not necessarily equal to $R$. However the discrepancy between these two rings can be dealt with when knowing enough parameters attached to $K$ such as the discriminant, etc. See e.g. the examples of quadratic and cubic fields given in Marcus' book "Number Fields", chap. 2. But there is no general systematic approach. Your example: $K=mathbf Q(sqrt 14), x= 4+sqrt 14, N(x)=2$. Here $R=mathbf Z(sqrt 14)$ because $14equiv 2$ mod $4$, so card $mathbf Z[sqrt 14]/(sqrt 14)=2$.

More generally, for a non zero ideal $I$ of $R$, the norm $N(I)$ can be defined as card $(R/I)$, but then, we must give an independent way to compute this cardinal. This is done by using the fact that $R$ is a Dedekind ring, in which any non zero $I$ can be be written uniquely as a product $I={P_1}^{m_1}...{P_r}^{m_r}$ of powers of prime ideals of $R$ (or maximal ideals, it's the same thing here). By the CRT, we are reduced to compute card $(R/P^m)$. For this, consider the chain of ideals $P^m<P^{m-1}<...<P<R$, in which two consecutive terms are of the form $J<PJ$, hence without any ideal strictly squeezed between them. It follows that, denoting by $p$ be the prime of $mathbf Z$ under $P$ and by $f_P$ the inertia index of $P$, the quotient $PJ/P$ can be viewed as a $1$-dimensional vector space over $R/P$, hence a vector space of dimension $f_P$ over $mathbf F_p$ . In other words, $[R/P^m : R/P]=mf_P$, i.e. card $(R/P^m)=p^{mf_P}$. Your example: $K=mathbf Q(sqrt 10), I=(2,sqrt 10)$. Here $R=mathbf Z[sqrt 10]$, again because $10equiv 2$ mod $4$. You could factorize $I$ and apply the above result, but it's quicker to consider the isomorphism of additive groups $R/Icong (R/mathbf Zsqrt 10)/ (I/mathbf Z sqrt 10)congmathbf Z/2mathbf Z$ (check).