Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to teach myself some notes on Local Fields, in advance for a course I will be taking next year. I covered some algebraic number theory a little while ago, but I'm having trouble understanding the notes I've been teaching myself with online; I was hoping someone could help clear up my confusions - I have a few and much of the material is totally new to me so I'm very grateful for your patience in advance!

The notes are located at - in particular, the proof on the bottom of page 16. The setup is as follows: $\mathfrak{p} \subset \mathcal{O}_K$ prime and $L / K$ a finite extension with $\mathfrak{p}\mathcal{O}_L = \mathcal{P}_1^{e_1}\ldots\mathcal{P}_r^{e_r},$ each $\mathcal{P}_i \subset \mathcal{O}_L$ prime, $f_i = [\mathcal{O}_L/\mathcal{P}_i : \mathcal{O}_K/\mathfrak{p}]$ (the "residue class degree"), then we want to prove that $\sum_{i=1}^r e_i\, f_i = [L:K]$.

Question 1: How do we actually know that $\mathcal{O}_L/\mathcal{P}_i$ is an extension of $\mathcal{O}_K/\mathfrak{p}$? I don't intiutively have a grasp of either of these, and since $\mathcal{P}_i$ divides $\mathfrak{p}$ in $\mathcal{O}_L$, it isn't obvious to me that $\mathcal{O}_L/\mathcal{P}_i$ should be 'bigger' than $\mathcal{O}_K/\mathfrak{p}$, let alone a field extension.

The first half of the proof aims to show that dim$_k (\mathcal{O}_L/\mathfrak{p}\mathcal{O}_L) = [L:K]$, where $k = \mathcal{O}_K/\mathfrak{p}$. To do so, we say:

i) By the Chinese Remainder Theorem, $\mathcal{O}_L / \mathfrak{p}\mathcal{O}_L \cong \mathcal{O}_L / \mathcal{P}_1^{e_1} \times \ldots \mathcal{O}_L/\mathcal{P}_r^{e_r}$. Any $x \in \mathcal{P}_i^a \backslash \mathcal{P}_i^{a+1}$ generates a quotient $\mathcal{P}_i^a / \mathcal{P}_i^{a+1}$ as an $\mathcal{O}_L$-module ("using properties of Dedekind domains")

Question 2: How do we know that $\mathcal{P}_i^a \backslash \mathcal{P}_i^{a+1} \neq \phi$? I'm fairly sure that this is something to do with the fact $\mathcal{P}_i$ is prime, but when I tried to prove it I couldn't get anywhere: is this correct? Also, is the deduction by "Use properties of Dedekind domains" particularly straightforward? It certainly wasn't obvious to me.

ii) Then dim$_{\mathcal{O}_L/\mathcal{P}_i}(\mathcal{P}_i^a/\mathcal{P}_i^{a+1})=1$ as a vector space (since generated by 1 vector). Then dim$_k(\mathcal{P}_i^a/\mathcal{P}_i^{a+1})=f_i$, and so dim$_k(\mathcal{O}_L/\mathcal{P}_i^{e_i})=e_i f_i$.

Question 3: How do we make this final deduction, that since dim$_k(\mathcal{P}_i^a/\mathcal{P}_i^{a+1})=f_i$, dim$_k(\mathcal{O}_L/\mathcal{P}_i^{e_i})=e_i f_i$? Are we using the tower law somehow? I can't see how this is deduced.

That's all my problems for now: sorry that this is such a lengthy "question", I'm aware that this isn't ideally suited to the single-problem nature of math.stackexchange, but I thought it would be a lot more efficient than setting up the same problem lots of times to ask similar questions. If there are any relevant books you can suggest which are simple and concise too then please don't hesitate to recommend.

share|cite|improve this question
I think it's a nice question: you listed your sources, included relevant excerpts and explained your issues. I think people here encourage unrelated questions to be split up, but this is all part of one proof so it seems fine to me. – Dylan Moreland Jan 19 '12 at 20:31
up vote 3 down vote accepted

Question 1. There is a natural embedding $\mathcal{O}_K\hookrightarrow \mathcal{O}_L$, and so you get a composite map $$\mathcal{O}_K\hookrightarrow \mathcal{O}_L\longrightarrow \mathcal{O}_L/\mathcal{P}_i.$$ What is the kernel of this composite map? It is $\mathcal{O}_K\cap\mathcal{P}_i$; but this intersection is precisely equal to $\mathfrak{p}$, since $\mathcal{P}_i$ lies above $\mathfrak{p}$; so the composite map induces an embedding $$\mathcal{O}_K/\mathfrak{p}\hookrightarrow \mathcal{O}_L/\mathcal{P}_i$$ by the isomorphism theorems. Since both are fields, this affords you the ability of viewing $\mathcal{O}_L/\mathcal{P}_i$ as a field extension of $\mathcal{O}_K/\mathfrak{p}$.

Question 2. $\mathcal{O}_L$ is a Dedekind domain, so it has unique factorization of ideals into products of prime ideals. In particular $I^{a+1}=I^a$ for an ideal $I$ and $a\gt 0$ if and only if $I=(0)$ or $I=\mathcal{O}_L$. Since $I^{a+1} = II^a\subseteq I^a$ for any ideal $I$, it follows that $\mathcal{P}_i^{a+1}\subset \mathcal{P}_i^a$, and the inclusion is proper.

Question 3. By basic properties of dimensions of vector spaces, $$\dim_k(\mathcal{O}_L/\mathcal{P_i}^{e_i}) = \dim_k(\mathcal{O}_L/\mathcal{P}_i) + \dim_k(\mathcal{P}_i/\mathcal{P}_i^2) + \dim_k(\mathcal{P}_i^2/\mathcal{P}_i^3) + \cdots + \dim_k(\mathcal{P}_k^{e_i-1}/\mathcal{P}_k^{e_i});$$ each summand on the right hand side is equal to $f_i$, and there are $e_i$ summands.

share|cite|improve this answer
Wow, that was very quick and extremely helpful, thankyou! One final thing I don't fully grasp is the statement 'any $x \in \mathcal{P}_i^a \backslash \mathcal{P}_i^{a+1}$ generates a quotient $\mathcal{P}_i^a / \mathcal{P}_i^{a+1}$ as an $\mathcal{O}_L$-module'. Is this result obvious given results on Dedekind domains, and am I just missing it, or is it nontrivial? Thanks again. – Warner Jan 19 '12 at 20:47
@Warner: The map from $\mathcal{O}_L/\mathcal{P}_i$ to $\mathcal{P}_i^a/\mathcal{P}_i^{a+1}$ given by sending $r$ to $xr$ is a (module) homomorphism, and induces a map $(x)\to \mathcal{P}_i^a/\mathcal{P}_i^{a+1}$, whose kernel is $(x)\cap\mathcal{P}_i^{a+1}$ and whose image is $((x)+\mathcal{P}_i^{a+1})/\mathcal{P}_i^{a+1}$. But $(x)+\mathcal{P}_i^{a+1}$ is the gcd of $(x)$ and $\mathcal{P}_i^{a+1}$, which is $\mathcal{P}_i^a$ by choice of $x$, and $(x)\cap\mathcal{P}_i^{a+1}$ is the lcm of $(x)$ and $\mathcal{P}_i^{a+1}$ which is $(x)\mathcal{P}_i$ (cont) – Arturo Magidin Jan 19 '12 at 21:10
@Warner: (cont); so $\mathcal{P}_i^a/\mathcal{P}_i^{a+1}$ is isomorphic to $(x)/(x)\mathcal{P}_i$, which is isomorphic to $\mathcal{O}_L/\mathcal{P}_i$, which is a 1-dimensional $\mathcal{O}_L$ module. We are using properties of Dedekind domains to identify $(x)\cap\mathcal{P}_i^{a+1}$ and $(x)+\mathcal{P}_i^{a+1}$. – Arturo Magidin Jan 19 '12 at 21:13
Perfect, thank you again for all the help! You certainly seem very on top of your algebraic number theory; I don't suppose you know of any books you ever found helpful at a fairly introductory level which I might locate, do you? – Warner Jan 19 '12 at 23:01
@Warner: I really like Daniel Marcus's Number Fields, from Universitext series, Springer-Verlag. It has lots of really nice exercises. – Arturo Magidin Jan 20 '12 at 1:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.