Integral closure of complete DVR in an algebraic extension of its fraction field Let $R$ be a complete discrete valuation ring of characteristic zero with residue field $k$ of positive characteristic $p$. Let $K=\mathrm{Frac}(R)$ and $L$ be an finite algebraic extension of $K$. Let $A$ be the integral closure of $R$ in $L$.
Questions. Why $A$ is also a discrete valuation ring and why completeness of $R$ imply that $A$ is finite over $R$ (i.e., "finite" as $R$-module)?
 A: Let $L = K(\alpha)$ and let $f(x)$ be the minimal polynomial of $\alpha$.
Fact: there is a bijection between non-zero prime ideals of $A$ and factors of $f(x)$ in $K_P[x]$, where $P$ is the unique prime ideal of $K$.
Note that $K=K_P$ since $K$ is a complete discrete valuation field.
Now since we can assume that $f(x)$ is irreducible, there exists exactly one prime ideal $Q$ in $A$. Hence $A$ is a DVR.
Lastly, $L_Q = K_P \cdot L = K \cdot L = L$, as required.
As to your second question: $B$ being a finitely generated $A$-module does not require completeness. It is a standard result proved using the trace function.
Edit: $P = Q \cap R$. Showing that there is such a $Q$ can be done by appealing to the Lying Over Theorem. The fact that $A$ is a DVR, i.e. that there is only one prime ideal in $A$, is harder to prove: the only way I know of involves what I quoted as a "fact". 
About $a_i$ lying in $A$: 
We prove that for $x \in A$, $\operatorname{tr}_{L/K}(x) \in R$. Let $M$ be the Galois closure of $L/K$. Since $x$ is integral over $R$, $\sigma_i(x)$ is integral too for all $\sigma_i \in \operatorname{Gal}(M/K)$. Hence $\sum \sigma_i(x) = \operatorname{tr}_{L/K}(x)$ is integral over $R$. As a trace (fixed by the Galois action), this must lie in $K$. Therefore $\operatorname{tr}_{L/K}(x)$ is an element of $K$ integral over $R$, and as $R$ is integrally closed, $\operatorname{tr}_{L/K}(x) \in R$.
