6
$\begingroup$

Let $n$ be an integer. There is a well-known formula for $\phi (n)$, where $\phi$ is the Euler phi function (totient). Essentially, $\phi(n)$ gives the number of invertible elements in $\mathbb{Z}/n\mathbb{Z}$.

Since Dedekind domains have the same factorization theorem for ideals analogous to that of the integers, one can define a generalized Euler phi function for an ideal $I$ of a Dedekind domain $R$, i.e. $\phi(I)$ gives the number of invertible elements of the factor ring $R/I$.

Moreover, $I$ factors uniquely into the product $$I=P_1^{r_1}P_2^{r_2}\ldots P_k^{r_k}$$ with $P_i$ prime, distinct. By the Chinese remainder theorem, $R/I$ is isomorphic to the product of the $k$ rings $R/P_k^{r_k}$. Therefore $\phi$ is multiplicative. Hence it suffices to consider the case $I=P^n$, $P$ prime, non-zero. Here my first problem: what is the number of invertible elements in $R/P^n$? Imitating the case $R=\mathbb{Z}$, it should be $\phi(P^n)=q^{n-1}(q-1)$, where $q$ is the cardinality of $R/P$, but i've not a proof of this.

Consider this example. Let $K=\mathbb{Q}[i]$ be a quadratic number field and $R=\mathscr{O}_k$ its ring of algebraic integers. We know that $\mathscr{O}_K=\mathbb{Z}[i]$. For every ideal $I$ of $\mathscr{O}_K$, put $\phi_K(I)$ equal to the number of invertible elements of the factor ring $\mathscr{O}_K/I$. I want to compute $$\phi(6630\mathscr{O}_K)$$ where $6630\mathscr{O}_K$ is the ideal generated by $6630$ in the extension ring $\mathscr{O}_K$ of $\mathbb{Z}$. This is what i did:

  • prime factorization: $6630=2\cdot3\cdot5\cdot13\cdot17$;

  • $2\mathscr{O}_K=(1+i)\mathscr{O}_K(1-i)\mathscr{O}_K$;

  • $3\mathscr{O}_K=3\mathscr{O}_K$;

  • $5\mathscr{O}_K=(1+2i)\mathscr{O}_K(1-2i)\mathscr{O}_K$;

  • $13\mathscr{O}_K=(3+2i)\mathscr{O}_K(3-2i)\mathscr{O}_K$;

  • $17\mathscr{O}_K=(1+4i)\mathscr{O}_K(1-4i)\mathscr{O}_K$

Hence $$\phi_K(6630\mathscr{O}_K)=\phi_K((1+i)\mathscr{O}_K)\phi_K((1-i)\mathscr{O}_K)\phi_K(3\mathscr{O}_K)\ldots\phi_K((1-4i)\mathscr{O}_K)$$ My problem is that i don't know how to compute cardinality of factor rings. For example, what is the cardinality of $\mathbb{Z}[i]/(1+i)\mathbb{Z}[i]$?..or the cardinality of $\mathbb{Z}[i]/(1+2i)\mathbb{Z}[i]$? The only fact i know is that all these factor rings are finite fields, hence of positive characteristic $p$, hence their cardinality is of the form $p^n$, but...what $p$, what $n$?

EDIT: ...the final result is $2^2\cdot 3\cdot 5^2\cdot 13^2\cdot 17^2=14652300$

$\endgroup$

1 Answer 1

3
$\begingroup$

Your question is very similar to this question.

What I didnt see there in a quick look is how to compute the cardinality of $R/P^n$:

If $n=1$ this is a finite field as you remarked, and its characteristic is exactly the rational prime lying under $P$ (which is $P\cap\mathbb{Z}$). you can see this since there is a natural embedding of fields $\mathbb{Z}/P\cap\mathbb{Z}\to R/P$. The degree of this extension is called the "inertia degree" and is usually denoted by the letter $f$. Hence the cardinality of $R/P$ is $p^f$. (To actually compute the inertia degree, you can read this to get a formula using the discriminant).

In general, $n$ arbitrary, we can compute the order of $R/P^n$ most easily by localizing at $P$ ($R/P\equiv R_P/P_P$) and there $P$ is generated by a single element $\pi$. There is the short exact sequence of $R_P$ modules: $$ 0\to (\pi^{n-1})/(\pi^n)\to R/P^n\to R/P^{n-1}\to 0$$ and since $(\pi^{n-1})/(\pi^n)$ is isomorphic to $R_P/(\pi)$ as an $R_P$ module, we get inductively the formula for the cardinality of $R/P^n$.

$\endgroup$
5
  • $\begingroup$ the only thing i don't understand are the last words of your answer: "so..is a vector space over .. with basis $1,\pi,\ldots\pi^{n-1}$ $\endgroup$
    – bateman
    Sep 8, 2013 at 9:27
  • $\begingroup$ oh.. I lied..this is not true at all, consider $\mathbb{Z}/4$ over $\mathbb{Z}/2$, it cannot be done $\endgroup$
    – edo arad
    Sep 8, 2013 at 9:45
  • $\begingroup$ fixed. sorry for being too hasty.. $\endgroup$
    – edo arad
    Sep 8, 2013 at 10:01
  • $\begingroup$ i realized i didn't completely get the last computation. Using the short exact sequence, it seems that the cardinality of $R/P^n$ is the $n$-th power of the cardinality of $R/P$....which is not what i expected $\endgroup$
    – bateman
    Sep 9, 2013 at 18:48
  • $\begingroup$ It is precisely the same as in $\mathbb{Z}/p^n\mathbb{Z}$. The norm of an ideal $I$ is defined by $N(I)=(\text{card}(R/I))$. It can be shown that the norm is multiplicative. Please ask if there is anything unclear about the computation itself. $\endgroup$
    – edo arad
    Sep 13, 2013 at 21:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .