# Isomorphism of quotient of algebraic number rings

Assume $L,K$ are two number fields and $\mathcal{O}_L,\mathcal{O}_K$ are their algebraic number rings. Let $p_L (p_K)$ be a prime ideal for $\mathcal{O}_L$ $(\mathcal{O}_K)$ respectively, $n,m >1$ are positive integers.

What if $\mathcal{O}_L/p_L^n \cong \mathcal{O}_K/p_K^m$ as commutative rings? Which invariants of $p_L,p_K$ will be the same? Are there non-trival isomorphic examples?

I am interested in this because quotients like $\mathcal{O}_L/p_L^n$ produce lots of examples of finite rings, so if they are not isomorphic we see examples of finite rings are rich.

Consider the residue degrees $f_{K,p_K}= [\mathcal O_K/p_K:\mathbb F_p]$ and $f_{L,p_L}=[\mathcal O_L / p_L:\mathbb F_q]$, $p,q$ rational primes.

Since $\mathcal O_L/p_L^n$, $\mathcal O_K/p_K^m$ are local (artinian) rings with residue fields $\mathcal O_L/p_L$, $\mathcal O_K/p_K$ resp. we have that $f_K = f_L$, $p=q$ and $m=n$, by cardinality.

An example (hope non-trivial), let $L=\mathbb Q(i)$, $K= \mathbb Q$. Then $p_L=(i+2)$ is a prime ideal of $\mathcal O_L= \mathbb Z[i]$ lying over $p_K=5$. This yields isomorphisms for every $m=n$.

Let $K$ a number field, $\mathfrak p$ a prime (and hence maximal) ideal of $\mathcal O_K$ and let $O_{\mathfrak p}\supseteq \mathbb Z_p$ denote the completion of $\mathcal O_K$ w.r.t. $\mathfrak p$. Since $O_\mathfrak p/\mathfrak p^nO_{\mathfrak p}\simeq \mathcal O_K/\mathfrak p^n$ for every $n$, the rings you consider appear naturally as quotients of finite extensions of $\mathbb Z_p$.

It is then clear that two number fields can have "the same" sequence of quotient rings $\{O/\mathfrak p^n\}_n$, only if their completions are isomorphic. The previous is an example of this fenomena since $\mathbb Z[i]_{(i+2)} \simeq \mathbb Z_5$. In particular, if the sequences of $K$ and $L$ are isomorphic then $$\begin{cases} e_{L,p_L} &= e_{K,p_K}\\ f_{L,p_L} &= f_{K,p_K} \end{cases}$$ i.e., ramification indexes and residue degrees coincide. Here $e_{K,p_K}$ is the maximal integer $e$ such that $p_K^e\mid p\mathcal O_K$, see this Wikipedia article.

The reciprocal is true if $e_{L,p_L}=1$ by local class field theory.

Related to that is the case $L=\mathbb Q(i)$, $K= \mathbb Q(\sqrt 2)$, $p_L=(i+1)$, $p_K= (\sqrt 2)$. We find isomorphism if and only if $m=n =1, 2\text{ or }3$. Maybe it is important to remark that isomorphism at level $n=m=k$ implies isomorphism at level $n=m=k-1$.

• Where $f_K$ is...? – user26857 Nov 13 '17 at 21:47
• Ups, sorry about that! – eduard Nov 14 '17 at 1:12