There is a fundamental equality in algebraic number theory:
Let $(A,\mathfrak{p})$ be a DVR (Discrete Valuation Ring), $K$ be its field of quotient, $L/K$ be a finite field extension of degree $[L:K] = n$ and $B$ is a subing of $K$ of field of quotient $L$ containing $A$. Assume $B$ is a finite $A$-module, then we know $B$ is a Dedekind domain. Let $\mathfrak{p}B =\mathfrak{P}_1^{e_1} \mathfrak{P}_2^{e_2} \cdot \cdot \cdot \mathfrak{P}_r^{e_r}$ and $f_i = [ B / \mathfrak{P}_i : A / \mathfrak{p} ]$, then we have $n = \sum_{i=1}^r e_i f_i$. (I have to assume that $B$ is integrally closed. If not, we still have such a formula but with $e_i$ being the length of some modules as below.)
Notice that we don't have to assume $L/K$ is separable if we assume $B$ is finite over $A$.
In general, if we only assume $B$ is integral over $A$ (not necessary finite), then we have $l_{A} (B / \mathfrak{p}B) = \sum_{i=1}^r e_i f_i$, here $l_A (B / \mathfrak{p}B)$ means the length of the $A$-module $B / \mathfrak{p}B$ and $e_i$ is the length of the $B_{\mathfrak{P}_i}$-module $B_{\mathfrak{P}_i} / \mathfrak{p} B_{\mathfrak{P}_i}$. We also have $n = [ L:K ]$ $\geq l_A (B / \mathfrak{p}B)$.
My question is that: Under the above assumption ($B$ is integral over $A$, not necessary finite over $A$), if we have $n = e_i f_i$, or equivalently, $[ L:K ]$ $= l_A (B / \mathfrak{p}B)$, we can conclude that $B$ is in fact finite over $A$?
In Kaplansky's book "Commutative rings" (Theorem 100), he gave an example of $L/K$ being a purely inseparable extension of degree 2 and $B$ and $A$ are DVRs such that $B$ is not finite over $A$. But in this example, $2 = [ L:K ]$ $> l_A (B / \mathfrak{p}B) = 1$. In order to find a counter-example for my guess, we have to consider another example than that one in Kaplansky's book, for which I can't figure out one. On the other hand, I can't give a proof for it neither.
