# A yet another theorem on the different ideal of algebraic number fields

I think I came up with a proof of the following theorem using non-archimedian completions. But I'm not 100% sure. Is this correct?

Theorem Let $A$ be a Dedekind domain, $K$ its field of fractions. Suppose that for each maximal ideal $P$ of $A$, the residue field $A/P$ is perfect. Let $L$ be a finite separable extension of $K$. Let $B$ be the integral closures of $A$ in $L$. Let $\alpha$ be an element of $B$. Let $f(X)$ be the characteristic polynomial of $\alpha$. Let $f'(X)$ be the formal derivative of $f(X)$. We denote $f'(\alpha)$ by $\delta(\alpha)$. Since $f(X) ∈ B[X]$, $\delta(\alpha) ∈ B$. Let $D_{L/K}$ be the different. Then $D_{L/K}$ is generated by the set $\{\delta(\alpha); \alpha ∈ B\}$.

Proof Let $I$ be the ideal generated by the set $\{\delta(\alpha); \alpha ∈ B\}$. 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 discrete valuation ring. 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 propositions, I think we are done.

Proposition 1 Let $K$, $L$, $A$, $B$, $D_{L/K}$ be as in the theorem. Let $I$ be the ideal generated by the set $\{\delta(\alpha); \alpha ∈ B\}$. Then $I$ $⊂$ $D_{L/K}$.

Proof Suppose that $\delta(\alpha)$ $≠$ 0. Let $f(X)$ be the characteristic polynomial of $\alpha$. Let $g(X)$ = $f(X)/(X - \alpha)$ = $X^{n-1}$ $+$ $a_{n-2}X^{n-2}$ $+ ... +$ $a_0$ $∈ B[X]$. Let $\mu ∈ (D_{L/K})^{-1}$. Let $h(X)$ = $Tr(\mu{f(X)}/(X - \alpha))$ = $Tr(\mu)X^{n-1}$  +  $Tr(\mu{a_{n-2}})X^{n-2}$  + ... +  $Tr(\mu{a_0})$. Since $\mu ∈ (D_{L/K})^{-1}$, $h(X) ∈ B[X]$. Replacing $X$ of $h(X)$ with $\alpha$, we get $\mu{f'(\alpha)}$ = $h(\alpha)$. Hence $\mu{f'(\alpha)} ∈ B$. Hence $f'(\alpha)$ $∈$ $D_{L/K}$. Therefore $I$ $⊂$ $D_{L/K}$.

Proposition 2 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.

Proposition 3 Let $A$ be a complete 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 $L$. 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 is immediate using proposition 2.

Proposition 4 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.

Proposition 5 Let $A$ be a Dedekind domain, $K$ its field of fractions. Let $L$ be a finite separable extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let P be the maximal ideal of A. Let Q be a maximal ideals of $B$ lying over $P$. Let $D_{L/K}$ be the different. Let $\hat{L_Q}$, $\hat{B_Q}$, $\hat{K_P}$ be completions. Let $(D_{L/K})_Q$ be the ideal of $\hat{B_Q}$ generated by $D_{L/K}$. Then $(D_{L/K})_Q$ is the different of $\hat{L_Q}$ over $\hat{K_P}$

The proof can be found in Serre's Local fields, chapter 3, section 4, Proposition 10 in page 52.

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