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 
$n = \sum_{i=1}^r e_i f_i$.

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 lenght 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, 
$\[ L:K \]$ $= l_A (B / \mathfrak{p}B)$, if 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.