Counterexample for the sum formula in Dedekind extension

Question

Let $A\subseteq B$ be Dedekind domains, and $K\subseteq L$ be their quotient fields respectively, and assume that $B$ is the integral closure of $A$ in $L$. Let $\mathfrak{p}$ be a prime ideal of $A$. If $$ \mathfrak{p}B=\prod_{i=1}^r\mathfrak{P}_i^{e_i} $$is the prime decomposition of $\mathfrak{p}B$ in $B$, in Serre's Local Fields it is proved the identity $$ \sum_{i=1}^r e_if_i=n $$where $n=[L:K]$ and $f_i=[B/\mathfrak{P}_i:A/\mathfrak{p}]$, under the hypothesis that $L/K$ is separable. He remarks that the sum of $e_if_i$ may $<n$ without the separable hypothesis. It would be appreciated if anyone can provide an example showing the case that $$ \sum_{i=1}^re_if_i<n $$can happen.

Answer 1

We need that $L/K$ is a finite extension and $B$ is the integral closure of $A$ in $L$, and that $A_\mathfrak{p} B$ is a finitely generated $A_\mathfrak{p}$-module (satisfied when $L/K$ is separable, by the trace map).

• The finitely generated hypothesis is needed to get that $A_\mathfrak{p} B$ is a free $A_\mathfrak{p}$-module of rank $[L:K]$ (as $A_\mathfrak{p}$ is a PID).

• A counter-example to the finitely generated hypothesis is given there p.200 and 212: take $f(t)\in \Bbb{F}_p[[t]],\not \in \overline{\Bbb{F}(t)}$, we get an embedding $\Bbb{F}_p[x,y]\to \Bbb{F}_p[t,f(t)]\subset \Bbb{F}_p[[t]]$ and a discrete valuation on the former. The image of $\Bbb{F}_p[x,y^p]$ is $\Bbb{F}_p[t,f(t^p)]$. The discrete valuation $v$ extends to the fraction fields and the obtained DVR $A=\{ a\in \Bbb{F}_p(x,y^p),v(a)\ge 0\},B=\{ b\in \Bbb{F}_p(x,y),v(b)\ge 0\}$ have $t$ as an uniformizer and the same residue field $\Bbb{F}_p$. Finally $B$ is clearly integral over $A$.

• If the finitely generated hypothesis is satisfied then $B/\mathfrak{p}\cong A_\mathfrak{p} B/\mathfrak{p}$ is a $[L:K]$-dimensional $A/\mathfrak{p}\cong A_\mathfrak{p}/\mathfrak{p}$ vector space.

  On the other hand by comaximality and unique factorization of ideals $$B/\mathfrak{p}= B/\prod_i\mathfrak{P}_i^{e_i}\cong \prod_i B/\mathfrak{P}_i^{e_i}$$ $B/\mathfrak{P}_i^{e_i}$ is a $e_if_i$-dimensional $A/\mathfrak{p}$ vector space.