A problem on Algebraic Number Theory, Norm of Ideals

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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