Vtyjkyuk

Let $A$ be a complete DVR with fraction field $K$ and residue field $k$, let $l$ be a finite separable extension of $k$, then $l=k(baralpha)$ for some $baralphain l$ whose minimal polynomial $bar gin k[x]$ is monic, irreducible and separable. Let $gin A[x]$ be any lift of $bar g$ that is monic, then $g$ is also irreducible and separable.

Define $L:= K[x]/(g(x))=K(alpha)$, where $alpha$ is the image of $x$ in $K[x]/(g(x))$, then $L$ has valuation ring $B$ which equals to $A[alpha]$.

Why is $B$, the valuation ring of $L = K(alpha)$, equal to $A[alpha]$?

I know since $A$ is a complete DVR, $B$ is the integral closure of $A$ is $L$, so $A[alpha]subset B$, but I don't know the other inclusion.

p.s. It is found in one step in the proof in a course manual, that if $A$ is a complete DVR with fraction field $K$ and residue field $k$, then there's a correspondence between unramified extensions $L/K$ and separable extensions $l,/,k$.

The reason is quite simple.

First we show $l$ is indeed the residual field $B/mathfrakP$, temporarily let $l'$ denote the residual field. Then $[l':k]leq [L:K]$. Also $$alpha in Limplies baralpha in l' implies l=k(baralpha) subset l'$$
Hence $[l':l]geq 1$. If $[l':l]>1$, then $[l:k] leq [L:K]/[l':l] < [L:K]$, contradicting to $[L:K]=[l:k]$. Therefore $l=l'$.

Let $n=[L:K]=[l:k]$, $mathfrakP$ be the prime ideal in $L$ and $v$ the valuation on $K$. Suppose $$k_0 + k_1 alpha + cdots + k_n-1 alpha^n-1 in B qquad k_iin K$$
If there is some $k_i$ such that $v(k_i)<0$, let this $i$ be such that $v(k_i)$ is minimal. Dividing both sides by $k_i$ gives
$$a_0 + a_1 alpha + cdots + a_n-1 alpha^n-1 in mathfrakP qquad v(a_j)geq 0,a_i=1$$
this says $1,baralpha,cdots,baralpha^n-1$ are linearly dependent over $k$, a contradiction.

Thanks for @pisco that done most of the work for me, though due to subtle detail of notations(like different definitions of $baralpha$), I will accept his and write my commentary here.

Let $hatalpha$ be the image of $alpha$ in $B/mathfrak P$, then by construction of $g$, $bar g(hatalpha) = 0$, since $bar g$ is irreducible by assumption, $hatalphaneq 0$ in $B/mathfrak P$, so $alphanotin mathfrak P$ hence $B=A[alpha]$.

p.s. We have $n=[k(hatalpha):k] leq [B/mathfrak P : k]leq [L:K] = n$, so $B/mathfrak P =k(hatalpha) approx k(baralpha) = l.$