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Let $K\subseteq L$ be number fields over the field of rationals $\Bbb Q$. with rings of integers $\mathcal{O}_K\subseteq \mathcal{O}_L$. Let $P$ be a prime ideal of $\mathcal{O}_L$, let $p$ be a prime ideal of $\mathcal{O}_K$, such that $P$ is over $p$.

The residue class degree $f$ is defined to be $f=[\mathcal{O}_L/P:\mathcal{O}_K/p]$. The norm of $P$ is the ideal $N(P)=p^f$. This is the usual definition of norm of an ideal. (See Serre's Local fields and Serge Lang's Algebraic tumber theory.)

Swinnerton-Dyer's A brief guide to algebraic number theory has a different definition of norm of an ideal (page 25). Namely if $A$ is an ideal of $\mathcal{O}_L$, it is defined as $N(A)$ = ideal in $\mathcal{O}_K$ generated by elements $N(a)$ where $a\in A$.

I don't 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?

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  • $\begingroup$ A very basic hint: The first norm you define is an integer; the second is an ideal. So they are not literally equal. One relation between them is that, when K is Q, the ideal is generated by the integer. $\endgroup$ Commented Dec 18, 2009 at 3:44
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    $\begingroup$ The first norm is also an ideal. Norm (P)=p^f where p is a prime ideal. Both definitions are ideals. $\endgroup$
    – 7-adic
    Commented Dec 18, 2009 at 4:03
  • $\begingroup$ Oh, I see. OK, forget that then. I seem to be making a lot of stupid mistakes tonight; maybe I should stop doing math for a bit. $\endgroup$ Commented Dec 18, 2009 at 5:14
  • $\begingroup$ It seems to me like this should follow easily from the fact that the norm is transitive in towers and that the residue degree is multiplicative in towers. I would look at the tower L > K > Q. $\endgroup$
    – user1884
    Commented Dec 18, 2009 at 16:36
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    $\begingroup$ @7-adic: it would be nice for future readers if the accepted answer was actually correct. Perhaps you could un-accept the incorrect answer and accept for example KConrad's. $\endgroup$
    – Alex B.
    Commented Mar 10, 2020 at 14:12

6 Answers 6

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[Edit: this answer is incomplete/incorrect, see rather this one]

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.

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    $\begingroup$ I am coming to this one somewhat late, but the solution is incorrect. First of all, the fact that the norm map on field extensions is transitive does not so easily imply the norm map as defined by Swinnerton-Dyer is transitive. The problem is that the definition commands you to take the norm of all elements in the ideal, and not all elements in the ideal of ${\mathcal O}_K$ generated by norms from an ideal in ${\mathcal O}_L$ are going to be norms from ${\mathcal O}_L$. You can't compute a norm by Sw-D's defn by using any generating set: the ideal $(1+2i,1-2i)$ in ${\mathbf Z}[i]$ (contd). $\endgroup$
    – KConrad
    Commented May 23, 2012 at 3:42
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    $\begingroup$ is (1), so it has norm (1) in ${\mathbf Z}$, but note the numbers $1+2i$ and $1-2i$ have norm 5, hence the ideal in ${\mathbf Z}$ generated by their norms is (5), which is not the norm of $(1+2i,1-2i)$. Although it's true that the norm map on ideals is transitive, it definitely requires real work to explain. The second error is that from the equation $p^{f({\mathfrak P}|p)} = {\rm N}_{K/{\mathbf Q}}N_{L/K}({\mathfrak P})$, we can't conclude that ${\rm N}_{L/K}({\mathfrak P})$ is some power of ${\mathfrak p}$: consider the equation $5 = {\rm N}_{{\mathbf Q}(i)/{\mathbf Q}}(\alpha)$. (contd.) $\endgroup$
    – KConrad
    Commented May 23, 2012 at 3:46
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    $\begingroup$ You can't tell me what $(\alpha)$ is for sure. Maybe it's $(1+2i)$, maybe it's $(1-2i)$. Knowing some ideal has a power of $p$ as its norm tells you that the ideal is a product of powers of primes lying over $p$, but that is all. You definitely can't reason just from the norm equation in the 2nd paragraph that ${\rm N}_{L/K}({\mathfrak P})$ is a power of $\mathfrak p$. Again, the conclusion is correct but the reasoning is incorrect (or, if I'm misunderstanding the reasoning, at the very least there is a very substantial gap). While the two definitions of ideal norm are equivalence (contd). $\endgroup$
    – KConrad
    Commented May 23, 2012 at 3:50
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    $\begingroup$ -- excuse me, are equivalent -- the proof of the equivalence is not as simple as the solution here suggests it is. $\endgroup$
    – KConrad
    Commented May 23, 2012 at 3:50
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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

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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.

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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*tf 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.

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A reference: Neukirch's book "Algebraic Number Theory", Chapter III, Proposition (1.6)(vi) on page 187.

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Inspired by KConrad's proof: Let $A$ be a Dedekind domain, $K=Frac(A)$, $L/K$ finite separable extension, $B$ the integral closure of $A$ in $L$. $I\subset B$ an ideal. Let's show $N_{B/A}(I)=\sum_{x\in I}N_{B/A}(x)$.

1.If it's true for $I_1,I_2$, then it's also true for $I_1I_2$: $$N(I_1I_2)=N(I_1)N(I_2)=(\sum_{x\in I_1}N(x))(\sum_{y\in I_2}N(y))=\sum_{x\in I_1,y\in I_2}N(xy)\subset\sum_{z\in I_1I_2}N(z).$$ So, we may assume $I=\mathfrak{P}$ a prime ideal.

2.Set $\mathfrak p=\mathfrak P\cap A$. Let $\mathfrak{P_1}=\mathfrak P,\mathfrak{P_2},\dots,\mathfrak{P_g}$ be the primes above $\mathfrak p$. We pick $a_1\in \mathfrak{P_1}-\mathfrak{P_1^2}$ and $a_j\in B-\mathfrak{P_j}$ for $j\ge2$. By Chinese Reminder Theorem, there's $x\in B$ such that: $x\equiv a_k(mod \mathfrak{P}_k)$ for any $k$. Then $x\in \mathfrak{P}$, $$v_{\mathfrak{p}}(N(x))=\sum_{k=1}^gv_{\mathfrak{P}_k}(x)f(\mathfrak{P}_k/\mathfrak{p})=v_{\mathfrak{p}}(N(I)).$$ For any $q$ a prime ideal of $A$ different from $\mathfrak{p}$, let $Q_1\dots,Q_h$ be the primes above $q$, again by CRT or argue directly, there's $y\in I$, such that $y\notin Q_j$ for any $j$. then $v_qN(y)=0=v_qN(I)$. so $N(I)=\sum_{x\in I}N(x)$.

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