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Theorem Let $K ⊂ L ⊂ E$ be a tower of algebraic number fields. Suppose that $E/K$ is a Galois extension. Let $B$ and $C$ be the rings of algebraic integers in $L$ and $E$ respectively. Let $G$ be the Galois group of $E/K$. Let $H$ be the Galois group of $E/L$. Let $G/H$ be the set of left cosets of $G$ by $H$. Let $S$ be the set of representatives of $G/H - \{H\}$. For each $\sigma ∈ S$, Let $J_\sigma$ be the ideal of $C$ generated by the set $\{x - \sigma(x); x ∈ B\}$. Let $J$ be $\prod_{\sigma ∈ S}J_\sigma$. Let $D_{L/K}$ be the different. Then $(D_{L/K})C = J$.

I know a proof of this theorem which uses a somewhat complicated argument using the Kronecker's method on forms. I'm looking for a simpler proof without forms. Perhaps one can prove this by using p-adic completions, but I prefer a proof using only ideals.

EDIT As a corollary to this theorem, we get the following useful result.

Corollary Let $L$ be a finite Galois extension of an algebraic number field K. Let $B$ be the ring of algebraic integers in $L$. Let $G$ be the Galois group of $L/K$. For each $\sigma ∈ G - \{1\}$, Let $J_\sigma$ be the ideal of $B$ genertated by the set $\{x - \sigma(x); x ∈ B\}$. Let $D_{L/K}$ be the different. Then $D_{L/K} = \prod_{\sigma ∈ G - \{1\}}J_\sigma$.

EDIT By the way, I guess the word the different came from the above theorem.

EDIT For the readers convenience, I will prove the theorem using the Kronecker's method. The proof is largely borrowed from the Takagi's book "Algebraic number theory"(in Japanese). I have simplified it using localizations. After all, it's not so complicated as I thought. For a technical reason, we prove the theorem under the following more general conditions. Let $A$ be a Dedekind domain, $K$ its field of fractions. Let $L$ be a finite separable extension of $K$. Let $E$ be a finite Galois extension of $K$ such that $L ⊂ E$. Let $B$ and $C$ be the integral closures of $A$ in $L$ and $E$ respectively.

Let $B[X_1, ..., X_m]$ be the polynomial ring over B. Let $F ∈ B[X_1, ..., X_m]$. We denote by $I(F)$ the ideal of $B$ generated by all the coefficients of $F$. Given $F, G ∈ B[X_1, ..., X_m]$, it can be proved by a simple argument which goes back to Gauss that $I(FG)$ = $I(F)I(G)$(for example, see theorem 13 in Hilbert's "The theory of algebraic number fields"). This fact is the basis of the Kronecker's theory of forms.

Since $B$ is a finitely generated $A$-module, there exist elements $\theta_1、...、\theta_m$ which generate $B$ as an $A$-module. Let $\eta$ = $\theta_1X_1 + ... + \theta_mX_m ∈ B[X_1, ..., X_m]$. Let $T = S ∪$ {1}. Let $G(Y)$ = $\prod_{\sigma ∈ T}(Y - \sigma(\eta)) ∈ B[X_1, ..., X_m][Y]$. Let $G'(Y)$ be the formal derivative of $G(Y)$ with respect to $Y$. Then $G'(\eta)$ = $\prod_{\sigma ∈ S}(\eta - \sigma(\eta)) ∈ B[X_1, ..., X_m]$. Let $I$ = $I(G'(\eta))$. Since $J_\sigma = CI(\eta - \sigma(\eta))$ for $\sigma ∈ S$, $J$ = $C(\prod_{\sigma ∈ S}I(\eta - \sigma(\eta)))$ = $CI(\prod_{\sigma ∈ S}(\eta - \sigma(\eta))) = CI$. It suffices to prove that $D_{L/K}$ = $I$.

For each prime ideal $P$ of $A$, let $(D_{L/K})_P$ and $I_P$ be the localizations of $D_{L/K}$ and $I$ with respect to the multiplicative subset $S = A - P$ respectively. It suffices to prove that $(D_{L/K})_P$ = $I_P$ for each prime ideal $P$ of $A$. Therefore, by replacing $A$ by $A_P$, we can assume that $A$ is a principal ideal domain. Since $B$ is finitely generated torsion-free A-module, $B$ is a free $A$-module. Let $\omega_1,...,\omega_n$ be its basis over $A$.

Let $\xi$ = $\omega_1X_1 + ... + \omega_nX_n ∈ B[X_1, ..., X_n]$. Let $F(Y)$ = $\prod_{\sigma ∈ T}(Y - \sigma(\xi)) ∈ B[X_1, ..., X_n][Y]$. Let $F'(Y)$ be the formal derivative of $F(Y)$ with respect to $Y$. Then $F'(\xi)$ = $\prod_{\sigma ∈ S}(\xi - \sigma(\xi)) ∈ B[X_1, ..., X_n]$. $N_{L/K}(I)$ = $N_{L/K}(I(F'(\xi)))$ = $I(N_{L/K}(F'(\xi)))$ = $I(\prod_{\sigma ∈ T}F'(\sigma(\xi))))$ = $I(\Delta[1, \xi, ..., {\xi}^{n-1}]^2)$, where $\Delta[1, \xi, ..., {\xi}^{n-1}]$ is the Vandermonde determinant(for the relation between norms and forms, see section 14 of the Hilbert's book).

We have $\xi^{i}$ = $\omega_1{u_{i,1}}$ $+ ... +$ $\omega_n{u_{i,n}}$, $i$ = $0, 1, ..., n-1$, where $u_{i,j} ∈ A[X_1, ..., X_n]$. Hence $\Delta[1, \xi, ..., {\xi}^{n-1}]$ = $\Delta[\omega_1 ..., \omega_n]det(U)$, where $\Delta[\omega_1 ..., \omega_n]$ is the determinant of a matrix $(\sigma(\omega_i))$ and $U$ is a matrix $(u_{i,j})$. Hence $\Delta[1, \xi, ..., {\xi}^{n-1}]^2$ = $\Delta[\omega_1 ..., \omega_n]^2(det(U))^2$.

Let $d_{L/K}$ be the discriminant of $L/K$. Since $(\Delta[\omega_1 ..., \omega_n]^2)A$ = $d_{L/K}$, $N_{L/K}(I)$ = $d_{L/K}I((det(U))^2)$. Therefore $N_{L/K}(I) ⊂ d_{L/K}$. We will show that $N_{L/K}(I)$ = $d_{L/K}$.

Let $\alpha$ be an element of $B$. Let $f(X)$ = $\prod_{\sigma ∈ T}(X - \sigma(\alpha))$. Let $f'(X)$ be the formal derivative of $f(X)$. We denote $f'(α)$ by $\delta(α)$. Since $f(X) ∈ B[X]$, $\delta(α) ∈ B$. Since $\delta(α)$ = $\prod_{\sigma ∈ S}(\alpha - \sigma(\delta))$, $\delta(\alpha) ∈ I$. It can be proved that $D_{L/K}$ is generated by the set $\{\delta(\alpha); \alpha ∈ B\}$. For a proof of this fact, see, for example, Lang's "Algebraic number theory" which uses non-Archemedian completions or Hecke's "Lectures on the theory of algebraic numbers" which uses only ideals. Hence $D_{L/K} ⊂ I$. Therefore there exists an ideal $M$ of $B$ such that $D_{L/K}$ = $IM$. Taking norms of the both sides, we get $N_{L/K}(D_{L/K})$ = $N_{L/K}(I)N_{L/K}(M)$. It is well-known that $N_{L/K}(D_{L/K})$ = $d_{L/K}$. Hence $d_{L/K} ⊂ N_{L/K}(I)$. Hence $N_{L/K}(I)$ = $d_{L/K}$ as desired. Then $N_{L/K}(M)$ = $A$ and $M$ = $B$. Therefore we get $D_{L/K}$ = $I$. QED

EDIT The following condition should be added in the above proof. For each maximal ideal $P$ of $A$, the residue field $A/P$ is perfect.

EDIT I think I found a proof of the theorem using P-adic completions. It is easy to see that the theorem holds if there exists $\alpha ∈ B$ such that $B$ = $A[\alpha]$. Combining the result and the following proposition, we are done.

Proposition Let $A$ be a discrete valuation ring. Let $K$ be its field of fractions. Let $L$ be a finite separable extension. Let $B$ be the integral closure of $A$ in $L$. Suppose that $B$ is a discrete valuation ring. Let $\bar{K}$ be the residue field of A. Let $\bar{L}$ be the residue field of B. Suppose that $\bar{L}$ is a separable extension of $\bar{K}$. Then there exists $\alpha ∈ B$ such that $B$ = $A[\alpha]$.

The proof can be found in Serre's Local fields, Proposition 12 in page 57, or Lang's Algebraic number theory, Proposition 3 in page 59.

EDIT In the above proof, we also need the following result.

Let $A$ be a discrete valuation ring, $P$ its maximal ideal. Let $\hat{A_P}$ be the completion of $A$ with respect to $P$-adic topology. Then $\hat{A_P}$ is faithfully flat over $A$.

The proof can be found in Bourbaki's Commutative Algebra, Chap. III $§$3, No.5, Prop. 9.

EDIT The question remains. How can we get rid of the use of P-adic completions from the proof?

EDIT I crossposted this question in MO.

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I'm probably just unfamiliar with field theory in general, but could you please give a reference or brief introduction to what $ D_{L/K} $ means? –  Eric Haengel May 12 '12 at 6:36
2  
@Rankeya No. It's not the discriminant of $L/K$. I wrote a link to the Wikipedia article on the different of $L/K$. –  Makoto Kato May 12 '12 at 8:54
    
@MakotoKato: Hmm, my bad. I am deleting my comment. –  Rankeya May 12 '12 at 11:02
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