This question refers to Jürgen Neukirch's book Algebraic Number Theory and to J.S. Milne's notes on Algebraic Number Theory. Specifically, to the proof of the fundamental identity (Neukirch: Chapter I, Proposition 8.2; Milne: Theorem 3.34).

Using Neukirch's notation throughout, both proofs show that $n = \dim_{\kappa}\mathcal{O}/\mathfrak{p} = \sum\limits_{i = 1}^{r}e_if_i$, where $\kappa = \mathcal{o}/\mathfrak{p}$, $L/K$ is a field extension of degree $n$ and $\mathfrak{p}\mathcal{O} = \mathfrak{P}_1^{e_1}...\mathfrak{P}_r^{e_r}$ is the factorization of $\mathfrak{p}$ in $\mathcal{O}$.

Though Neukirch only proves it for separable extensions, leaving the general case to a later chapter.

My question is about the proof that $n = \dim_{\kappa}\mathcal{O}/\mathfrak{p}$. Here Neukirch's and Milne's proofs differ.

Milne first considers the case where $\mathcal{O}$ is a free $\mathcal{o}$-module, and tensors the module isomorphism $\mathcal{O} = \mathcal{o}^m$ with $K$ and with $\mathcal{o}/\mathfrak{p}$ to show that $m = n$ and that $\mathcal{O}/\mathfrak{p}\mathcal{O} = (\mathcal{o}/\mathfrak{p})^m$, respectively. He then proves the general case using some sort of localization argument (which I did not quite understand, by the way) and the fact that $\mathcal{o}_\mathfrak{p}$ is a PID.

On the other hand, Neukirch shows that a basis for $\mathcal{O}/\mathfrak{p}\mathcal{O}$ over $\kappa$ lifts to a basis for $L$ over $K$. He proves linear independence by choosing a particular element of $\mathcal{o}$ which, when multiplied by a nontrivial linear dependence relation over $\mathcal{o}$, gives rise to another dependence relation over $\mathcal{o}$ where not all coefficients lie in $\mathfrak{p}$, contradicting the linear independence over $\mathcal{o}/\mathfrak{p}$. He proves that the proposed basis spans $L$ by using a Nakayama's lemma type of argument (using separability to show that $\mathcal{O}$ is a finitely-generated $\mathcal{o}$-module).

On the surface, the two proofs look quite different to me, but I was wondering if they aren't really just two ways of looking at the same thing. In particular it seems to me like Milne's proof is slicker, while Neukirch uses a more explicit argument. Does the tensor product hide some of the details in Neukirch's proof?

  • $\begingroup$ (If someone can show me how to type the lowercase calligraphic o in LaTex I would also be grateful) $\endgroup$
    – Tob Ernack
    May 2, 2017 at 18:50

1 Answer 1


They are certainly not the same proof. In fact, this is a very good illustration of how localization and completion are very important techniques in algebraic number theory to simplify the situation when trying to prove an assertion which is trivially true for a specific class of "well behaved" cases.

Milne's idea is this: if $ \mathcal O $ is a free $ o $-module of finite rank $ n $, then clearly $ \mathcal O / \mathfrak p \mathcal O \cong ( o/\mathfrak p)^n $, on the other hand, the Chinese remainder theorem gives

$$ \mathcal O/\mathfrak p \mathcal O \cong \prod_{k=1}^m \mathcal O/\mathfrak q_k^{e_k} $$

where $ \mathfrak p \mathcal O = \mathfrak q_1^{e_1} \mathfrak q_2^{e_2} \ldots \mathfrak q_m^{e_m} $ is the prime factorization. Then, the result follows simply by dimension counting on both sides, over the field $ o/\mathfrak p $.

Now, note that the rank $ n $ is the degree of the field extension $ L/K $, thus given a not necessarily free $ \mathcal O $, we may localize both $ o $ and $ \mathcal O $ at the subset $ S = o - \mathfrak p $. This means that $ o $ becomes a principal ideal domain, and $ \mathcal O $ is a finitely generated torsion-free $ o $-module, thus it is free. Localization does not change the degree $ n = [L : K] $ (since it does not change the field of fractions of either ring), and it does not change the way that $ \mathfrak p $ splits in the ring extension. Therefore, we deduce the general result.

Neukirch, on the other hand, uses an argument which is very reminiscent of the way that the fundamental identity is derived in the local case, when we are dealing with general valued fields. I will not outline that argument here, but it is Proposition 6.8 in Chapter 2 of Neukirch's book. Reading that proof may give you some insight on how Neukirch came up with the proof in the number field case - he is simply using the same argument and adapting it so that it works without completions.

  • $\begingroup$ Thank you for your answer! Is it trivial to see that the factorization does not change in the localized ring (and that the residue class degrees of the factors remain the same)? $\endgroup$
    – Tob Ernack
    May 2, 2017 at 21:19
  • $\begingroup$ @TobErnack You may simply write out the same factorization and use the fact that primes of $ S^{-1} R $ correspond bijectively with prime ideals of $ R $ not meeting $ S $. For the residue class degrees, one notes that if $ I $ is a maximal ideal of a domain $ R $ such that $ I \cap S = \emptyset $, then any element $ r/s \in S^{-1} R $ is congruent to an element in the image of the embedding $ R \to S^{-1} R $. Thus, the obvious map $ R/I \to S^{-1} R/S^{-1} I $ is an isomorphism, i.e the cardinalities of the residue fields are unchanged by localization. $\endgroup$
    – Ege Erdil
    May 2, 2017 at 21:31
  • $\begingroup$ ...Then clearly $\mathcal O / \mathfrak p \mathcal O \cong ( o/\mathfrak p)^n$... How can I see this? $\endgroup$ Jan 19 at 21:27

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