1

Let $R$ be a ring of algebraic integers, $p ∈\mathbb{Z}$ a prime integer. Then the set $\mathcal A$ of all prime ideals $P ⊂ R$ such that $P ∩ \mathbb{Z}= p\mathbb{Z}$ is finite and nonempty. Also, $pR = ∏P^{e_P}$ for some positive integers $e_P$, where the product is taken over all $P ∈\mathcal A$. Let $n$ be the degree of the quotient field of $R$ over $\mathbb{Q}$. Show that for every $P ∈\mathcal A$, the factor ring $R/P$ is a finite field with $p^{f_P}$ elements for some integer $f_P$. Prove that $n = ∑ e_P f_P$ , where the sum is taken over all $P ∈\mathcal A$. (Hint: Compute the number of elements in $R/pR$.)

I have shown that $R/P$ is a finite field, and by Chinese Remainder Theorem, $R/pR \cong \bigoplus R/P^{e_P}$. Also, since $p$ is in $\mathbb{Q}$, $N(p)=p^n$. It remains to show that $|R/pR|=p^n$ and $|R/P^{e_P}|=|R/P|^{e_P}$. But I don't know how to relate these together.

user26857
  • 52,094

1 Answers1

1

$|R/pR| = p^n$ can be shown like this: It is well known that $R \cong \mathbb Z^n$ as abelian groups. We have a commutative diagram of abelian groups

$$\require{AMScd} \begin{CD} R @>{\cong}>> \mathbb Z^n\\ @VV{\cdot p}V @VV{\cdot p}V \\ R @>{\cong}>> \mathbb Z^n \end{CD}$$

, hence the co-kernels of the two vertical maps are isomorphic as abelian groups, i.e. $|R/pR|=|\mathbb Z^n/p\mathbb Z^n|=p^n$.


We are left to show $|R/P^e| =|R/P|^e$. We have the exact sequence

$$0 \to P^e/P^{e+1} \to R/P^{e+1} \to R/P^{e} \to 0,$$

hence it suffices to show $|P^e/P^{e+1}| = |R/P|$. Then the sequence inductively proves the statement.

The equality follows from the following more general result about Dedekind Domains.

MooS
  • 31,390
  • There is no need to invoke a general result for Dedekind domains: just localize the exact sequence at $P$ and use that $R_P$ is a DVR. – user26857 Apr 20 '16 at 09:12