Why $B\otimes_A \hat A\backsimeq \prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q$ if they are isomorphic after reducing modulo $\hat{\mathfrak p}$?

Question

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 $B\otimes_A \hat A\backsimeq \prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q$.

The proof said:

Since $A$ is $DVR$, $B$ is the free $A$-module of rank $n=[L:K]$, and therefore $B\otimes_A \hat A$ is a free $\hat A$-module of rank $n$, and $\prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q$ is a free $\hat A$-module of rank $\sum e_{\mathfrak q}f_{\mathfrak q} = n$. These two $\hat A$-modules lie in isomorphic $K_{\mathfrak p}$- vector spaces $L\otimes_K K_{\mathfrak p}\backsimeq \prod L_{\mathfrak q}.$ ($K_\mathfrak p$ denotes the completion of $K$ here.)

Now to show $B\otimes_A \hat A\backsimeq \prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q$, it suffices to check they are isomorphic after reducing modulo $\hat {\mathfrak p}$, the maximal ideal of $\hat A$. That is, to check $B\otimes_A \hat A/\hat{\mathfrak p}\backsimeq \prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q/\hat{\mathfrak p}\hat B_\mathfrak q$.

Why it is suffice to show they are isomorphic after reducing modulo $\hat{\mathfrak p}$? 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.)

Answer 1

By Nakayama's lemma, it's known that

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

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

Now let $M=B\otimes_A \hat A, N=\prod_{\mathfrak q|\mathfrak p}\hat B_\mathfrak q$ in lemma 1, the constructed $\phi:M\rightarrow N$ by assumption induced an isomorphism $M/\hat {\mathfrak p}M \backsimeq N/\hat{\mathfrak p}N$, so $\phi:M\rightarrow N$ is a surjection.

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