Vtyjkyuk

I had read a Corollary:

Assume usual $AKLB$ conditions($L/K$ finite separable field extension and $B$ the integral closure of $A$ in $L$.) with $A$ a $DVR$ with maximal ideal $mathfrak p$. Let $mathfrak pB=prod mathfrak q^e_q$ be the factorization of $mathfrak pB$ in $B$. Let $hat A$ denote the completion of $A$, and for each $mathfrak q|mathfrak p$, let $hat B_mathfrak q$ denote the completion of $B_mathfrak q$. ($B_mathfrak q$ is the localization of $B$ at $mathfrak q$.)

Then $Botimes_A hat Abacksimeq prod_mathfrak qhat B_mathfrak q$.

The proof said:

Since $A$ is $DVR$, $B$ is the free $A$-module of rank $n=[L:K]$, and therefore $Botimes_A hat A$ is a free $hat A$-module of rank $n$, and $prod_mathfrak qhat B_mathfrak q$ is a free $hat A$-module of rank $sum e_mathfrak qf_mathfrak q = n$. These two $hat A$-modules lie in isomorphic $K_mathfrak p$- vector spaces $Lotimes_K K_mathfrak pbacksimeq prod L_mathfrak q.$ ($K_mathfrak p$ denotes the completion of $K$ here.)

Now to show $Botimes_A hat Abacksimeq prod_mathfrak qhat B_mathfrak q$, it suffices to check they are isomorphic after reducing modulo $hat mathfrak p$, the maximal ideal of $hat A$. That is, to check $Botimes_A hat A/hatmathfrak pbacksimeq prod_mathfrak qhat B_mathfrak q/hatmathfrak phat B_mathfrak q$.

Why it is suffice to show they are isomorphic after reducing modulo $hatmathfrak p$? Any other direct proof is welcome and I would also accept as an answer.

(Though the corollary doesn't specify what isomorphism it is, I think it is as an $hat A$-algebra isomorphism.)

By Nakayama's lemma, it's known that

Lemma 1. Let $A$ a local ring with maximal ideal $m$, $phi: Mrightarrow N$ a homomorphism of finitely generated $A$-modules, then the surjectivity of $barphi:$ $M/mM rightarrow N/mN$ implies $phi$ surjective.

Lemma 2. Let $M$ finitely generated $A$-module, if $phi:Mrightarrow M$ is surjective, then $phi$ is also injective.

Now let $M=Botimes_A hat A, N=prod_mathfrak qhat B_mathfrak q$ in lemma 1, the constructed $phi:Mrightarrow N$ by assumption induced an isomorphism $M/hat mathfrak pM backsimeq N/hatmathfrak pN$, so $phi:Mrightarrow N$ is a surjection.

Furthermore, since they are both isomorphic to free $hat A$-module $hat A^n$, we get a surjection $psi: hat A^n backsimeq Mrightarrow N backsimeq hat A^n$ induced from $phi$ above, so by lemma 2, $psi$, hence $phi$ is injective, that is, we get $phi$ is an isomorphism.