Let $L/K$ be an extension of number fields. Suppose $\theta\in \mathcal{O}_L$ is a primitive element of this extension with $f(X)\in\mathcal{O}_K[X]$ its minimal polynomial over $K$. Let $\mathfrak{p}$ be a (nonzero) prime ideal of $\mathcal{O}_K$. Let $\bar{f}(X)=f(X)\bmod \mathfrak{p}\in (\mathcal{O}_K/\mathfrak{p})[X]$ be the reduced polynomial over the finite field $\mathcal{O}_K/\mathfrak{p}$.

A classical theorem of Dedekind states that under some technical condition, if $\bar{f}$ factorizes into irreducible factors $$ \bar{f}=\prod_{i=1}^k \bar{f}_i^{m_i} $$ over $\mathcal{O}_K/\mathfrak{p}$, where $f_i$ are distinct and $\deg(f_i)=d_i$. Then $\mathfrak{p}\mathcal{O}_L$ splits into prime ideals $$ \mathfrak{p}\mathcal{O}_L=\prod_{i=1}^k \mathfrak{P}_i^{e(\mathfrak{P}_i/\mathfrak{p})} $$ with the ramification indices $e(\mathfrak{P}_i/\mathfrak{p})=m_i$ and the inertia degrees $f(\mathfrak{P}_i/\mathfrak{p}):=[\mathcal{O}_L/\mathfrak{P}_i: \mathcal{O}_K/\mathfrak{p}_i]=d_i$. Furthermore $\mathfrak{P}_i=\mathfrak{p}\mathcal{O}_L+f_i(\theta)\mathcal{O}_L$ where $f_i(X)\in\mathcal{O}_K[X]$ lifts $\bar{f}_i(X)$.

The technical condition is used to guarantee that the map $\mathcal{O}_K[\theta]/\mathfrak{p}\mathcal{O}_K[\theta]\to \mathcal{O}_L/\mathfrak{p}\mathcal{O}_L$ induced from the inclusion $\mathcal{O}_K[\theta]\hookrightarrow \mathcal{O}_L$ is an isomorphism. See Proposition (8.3), Chapter 1 in Neukirch's Algebraic Number Theory. Also see this MO post and Keith Conrad's note.

My question is what we can say when the map $\mathcal{O}_K[\theta]/\mathfrak{p}\mathcal{O}_K[\theta]\to\mathcal{O}_L/\mathfrak{p}\mathcal{O}_L$ is not an isomoprhism.

For an irreducible factor $\bar{f}_i$ of $\bar{f}$, the ideal $\mathfrak{p}_i:=\mathfrak{p}\mathcal{O}_K[\theta]+f_i(\theta)\mathcal{O}_K[\theta]$ of $\mathcal{O}_K[\theta]$ is a prime (and maximal) ideal of $\mathcal{O}_K[\theta]$ since $$ \mathcal{O}_K[\theta]/\mathfrak{p}_i\cong (\mathcal{O}_K/\mathfrak{p})[X]/(\bar{f}_i(X)) $$ is an extension of $\mathcal{O}_K/\mathfrak{p}$. Indeed it is easy to see that $\mathfrak{p}_1,\dots,\mathfrak{p}_k$ are precisely the prime ideals of $\mathcal{O}_K[\theta]$ that contains $\mathfrak{p}$. Then for every prime ideal $\mathfrak{P}$ of $\mathcal{O}_L$ lying over $p$, the ideal $\mathfrak{P}\cap \mathcal{O}_K[\theta]$ is some $\mathfrak{p}_i$. Denote by $I_i$ the set of prime ideals $\mathfrak{P}$ of $\mathcal{O}_L$ lying over $\mathfrak{p}$ satisfying $\mathfrak{P}\cap \mathcal{O}_K[\theta]=\mathfrak{p}_i$.

It is easy to see that $d_i$ divides $f(\mathfrak{P}/\mathfrak{p})$ for all $\mathfrak{P}\in I_i$ since $d_i=[\mathcal{O}_K[\theta]/\mathfrak{p}_i: \mathcal{O}_K/\mathfrak{p}]$ and the map $\mathcal{O}_K[\theta]/\mathfrak{p}_i\to \mathcal{O}_L/\mathfrak{P}$ is injective. What else can we say?

It seems to me that we should still have $$ m_i d_i=\sum_{\mathfrak{P}\in I_i} e(\mathfrak{P}/\mathfrak{p})f(\mathfrak{P}/\mathfrak{p}) $$ Is this true? Is there a reference for such a result? Thanks.

(Originally I conjectured $m_i=\sum_{\mathfrak{P}\in I_i} e(\mathfrak{P}/\mathfrak{p})$ but it is apparently false, as can be seen by considering the splitting of $2$ in $\mathbb{Q}(\sqrt{5})$.)

  • $\begingroup$ How does what you said compare to example 2 in Conrad's note? A primitive element for $\mathbf{Q}(\sqrt 5)$ is certainly $\sqrt 5$, with minimal polynomial $X^2 -5$, but mod 2 this is $(X+1)^2$. So what will you mean by "extend" or "inertia degree" in this context? $\endgroup$ – stankewicz Aug 22 '16 at 9:13
  • $\begingroup$ @stankewicz I see. I think it is a counterexample to what I said. Then is it still true that $\sum_{\mathfrak{P}\in I_i} e(\mathfrak{P}/\mathfrak{p})f(\mathfrak{P}/\mathfrak{p})=m_i d_i$? In this example $m=2$, $d=1$ for the factor $X+1$. And $e(\mathfrak{P}/\mathfrak{p})=1$, $f(\mathfrak{P}/\mathfrak{p})=2$ where $\mathfrak{P}$ and $\mathfrak{p}$ are both generated by 2 and are prime. $\endgroup$ – Zeyu Aug 22 '16 at 9:25
  • $\begingroup$ @stankewicz In that example, $\mathfrak{P}=(2)$ extends the ideal generated by $2$ and $\theta+1$ since $\theta+1\in (2)$. This follows from the facts that $(\theta+1)^2\in (2)$ and $(2)$ is prime. $\endgroup$ – Zeyu Aug 22 '16 at 9:45
  • $\begingroup$ Ah, but then shouldn't your logic also apply to $\sqrt{17}$? If you want to come up with a concept like "inertia degree" or "extends" in the case of a non-maximal order, you're going to have to write down a rigorous definition at some point and with one of those you might have some trouble. $\endgroup$ – stankewicz Aug 22 '16 at 11:23
  • $\begingroup$ @stankewicz I've modified the question and made it more rigorous. By extending I really mean containing. The inertia degrees and ramification indices are still defined for primes of $\mathcal{O}_L$ and I am trying to relate them to the degrees $d_i$ and multiplicities $m_i$ of the irreducible factors of $\bar{f}$. $\endgroup$ – Zeyu Aug 22 '16 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.