A problem on Algebraic Number Theory, Norm of Ideals - MathOverflow

A problem on Algebraic Number Theory, Norm of Ideals

2009-12-18T03:13:07Z

A problem on Algebraic Number Theory

K and L are number fields over Q.(Q is rational number filed) K is a subfield of L.

O_K is the integers of K. and O_L is the integers of L.

P is a prime ideal of O_L. p is a prime ideal of O_K. P is over p.

The residue class degree f is defined to be f=[O_L/P:O_K/p]. The norm of P is Norm(P)=p^f

This is the usual definition of Norm of an ideal.(See Serre's Local fields and Serge Lang's Algebraic Number Theory)

Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory has a different definition on Norm of an ideal.(Page 25)

if A is an ideal of O_L, Norm(A)= ideal in O_K generated by elements Norm(a) where a is in A.

I dont know why these two definitions are the same. Swinnerton-Dyer claims so in his book. Can anyone here give a hint, an explanation or anything else?

Answer by Rado for A problem on Algebraic Number Theory, Norm of Ideals

2009-12-18T07:16:46Z

I thought I should know this but then i ended up looking it up. Its in Lang's Algebraic Nubmer Theory pages 24-26 (at least for A principal, but that should be enough). That is if you want to know the proof. I have no idea where the intuition comes from but I bet it is using lattices somehow.

Answer by H. Hasson for A problem on Algebraic Number Theory, Norm of Ideals

2009-12-18T07:40:01Z

Hmm... I can see offhand how to deal with it if L/K is Galois, but I'd have to think about it otherwise... In the Galois case, above p you have r many prime ideals, each with ramification index e, and residue degree f. The rough sketch is to view this as a problem about discrete valuations, rather than prime ideals.

N(P) (according to your second definition) = < N(a)| a in P >. We know this is an ideal in O_K, and it only remains to describe its decomposition into primes. Since the ramification index of p in (each) P above it is e, the minimal p-adic valuation of an element in N(P) is f. So if t is a parametrizing element of the p-adic valuation (choose it in O_K), then u*t^f generates N(P)_p where u is in O_K - P (check that N(P) isn't divisible by other prime ideals, with similar methods).

Hope that helps a bit with the intuition.

After reading Adam's solution, I noticed a few things were wrong in my argument. They were corrected in the body.

Answer by Adam Topaz for A problem on Algebraic Number Theory, Norm of Ideals

2009-12-18T16:48:25Z

Ok, here's the argument: First recall that the usual norm for non-zero elements of a field is transitive in towers; thus the same is true for your second definition of the norm of an ideal. In particular, $N_{K|Q}\circ N_{L|K} = N_{L|Q}$. The fact that the norm $N_{L|Q}(\mathfrak{P}) = [\mathcal{O}_L:\mathfrak{P}] \cdot \mathbb{Z}$ is easy to see for a prime $\mathfrak{P}$ in $\mathcal{O}_L$; edit: and thus the same is true for any integral ideal $\mathfrak{a}$. Now let $\mathfrak{p} = \mathcal{O}_K \cap \mathfrak{P}$ and $(p) = \mathbb{Z} \cap \mathfrak{P}$.

We have $N_{L|Q}(\mathfrak{P}) = p^{f(\mathfrak{P}|p)} = N_{K|Q}N_{L|K}\mathfrak{P}$. In particular, we deduce that $N_{L|K}\mathfrak{P} = \mathfrak{p}^d$ for some $d$. Moreover, we know that

$N_{K|Q}\mathfrak{p}^d = p^{d \cdot f(\mathfrak{p}|p)} = p^{f(\mathfrak{P}|p)}.$

But then $d = f(\mathfrak{P}|p) / f(\mathfrak{p}|p) = f(\mathfrak{P}|\mathfrak{p})$ as required.

Answer by KConrad for A problem on Algebraic Number Theory, Norm of Ideals

2012-05-23T04:42:36Z

Here is a proof that the ideal norm as defined in the books by Serre and Lang is equal to the ideal norm as defined in Swinnerton-Dyer's book. We will start from the definition given by Serre and Lang, state some of its properties, and use those to derive the formula as given by Swinnerton-Dyer.

Background: Let $A$ be a Dedekind domain with fraction field $K$, $L/K$ be a finite separable extension, and $B$ be the integral closure of $A$ in $L$. For any prime $\mathfrak P$ in $B$ we define ${\rm N}_{B/A}({\mathfrak P}) = \mathfrak p^f$, where $f = f({\mathfrak P}|{\mathfrak p})$ is the residue field degree of $\mathfrak P$ over $\mathfrak p$, and this norm function is extended to all nonzero ideals of $B$ by multiplicativity from its definition on (nonzero) primes in $B$.

Properties.

1) The map ${\rm N}_{B/A}$ is multiplcative (immediate from its definition).

2) Good behavior under localization: for any (nonzero) prime ${\mathfrak p}$ in $A$, ${\rm N}_{B/A}({\mathfrak b})A_{\mathfrak p} = {\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}({\mathfrak b}B_{\mathfrak p})$ . Note that $A_{\mathfrak p}$ is a PID and $B_{\mathfrak p}$ is its integral closure in $L$; the ideal norm on the right side is defined by the definition above for Dedekind domains, but it's more easily computable because $B_{\mathfrak p}$ is a finite free $A_{\mathfrak p}$-module on account of $A_{\mathfrak p}$ being a PID and $L/K$ being separable. The proof of this good behavior under localization is omitted, but you should find it in books like those by Serre or Lang.

3) For nonzero $\beta$ in $B$, ${\rm N}_{B/A}(\beta{B}) = {\rm N}_{L/K}(\beta)A$ , where the norm of $\beta$ on the right is the field-theoretic norm (determinant of multiplication by $\beta$ as a $K$-linear map on $L$). To prove this formula, it is enough to check both sides localize the same way for all (nonzero) primes $\mathfrak p$: ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\beta{B}_{\mathfrak p}) = N_{L/K}(\beta)A_{\mathfrak p}$ for all $\mathfrak p$. If you know how to prove over the integers that $[{\mathcal O}_F:\alpha{\mathcal O}_F] = |{\rm N}_{F/{\mathbf Q}}(\alpha)|$ for any number field $F$ then I hope the method you know can be adapted to the case of $B_{\mathfrak p}/A_{\mathfrak p}$, replacing ${\mathbf Z}$ with the PID $A_{\mathfrak p}$. That is all I have time to say now about explaining the equality after localizing.

Now we are ready to show ${\rm N}_{B/A}({\mathfrak b})$ equals the ideal in $A$ generated by all numbers ${\rm N}_{E/F}(\beta)$ as $\beta$ runs over $\mathfrak b$.

For any $\beta \in \mathfrak b$, we have $\beta{B} \subset \mathfrak b$, so ${\mathfrak b}|\beta{B}$. Since ${\rm N}_{B/A}$ is multiplicative, ${\rm N}_{B/A}({\mathfrak b})|{\rm N}_{E/F}(\beta)A$ as ideals in $A$. In particular, ${\rm N}_{E/F}(\beta) \in {\rm N}_{B/A}({\mathfrak b})$ . Let $\mathfrak a$ be the ideal in $A$ generated by all numbers ${\rm N}_{E/F}(\beta)$ , so we have shown $\mathfrak a \subset {\rm N}_{B/A}(\mathfrak b)$ , or equivalently ${\rm N}_{B/A}(\mathfrak b)|\mathfrak a$ . To prove this divisibility is an equality, pick any prime power ${\mathfrak p}^k$ dividing $\mathfrak a$. We will show ${\mathfrak p}^k$ divides ${\rm N}_{B/A}(\mathfrak b)$ .

To prove ${\mathfrak p}^k$ divides ${\rm N}_{B/A}(\mathfrak b)$ when ${\mathfrak p}^k$ divides $\mathfrak a$, it suffices to look in the localization of $A$ at $\mathfrak p$ and prove ${\mathfrak p}^kA_{\mathfrak p}$ divides ${\rm N}_{B/A}(\mathfrak b)A_{\mathfrak p}$ , which by the 2nd property of ideal norms is equal to ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(\mathfrak b{B_{\mathfrak p}})$ . Since $B_{\mathfrak p}$ is a PID, the ideal ${\mathfrak b}B_{\mathfrak p}$ is principal: let $x$ be a generator, and we can choose $x$ to come from $\mathfrak b$ itself. By the 3rd property of ideal norms, ${\rm N}_{B_{\mathfrak p}/A_{\mathfrak p}}(xB_{\mathfrak p}) = {\rm N}_{E/F}(x)A_{\mathfrak p}$ . Showing ${\mathfrak p}^kA_{\mathfrak p}$ divides ${\rm N}_{E/F}(x)A_{\mathfrak p}$ is the same as showing ${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$ . Since $x$ is in in $\mathfrak b$, ${\rm N}_{E/F}(x) \in \mathfrak a \subset {\mathfrak p}^k$ , so ${\rm N}_{E/F}(x) \in {\mathfrak p}^kA_{\mathfrak p}$ . QED