I can't understand a detail in the proof of theorem 1.3.16 in Liu. The theorem is: let $(A,\mathfrak{m})$ a Noetherian local ring, $\hat{A}$ its $\mathfrak{m}$-adic completion, $(B,\mathfrak{n})$ an other local ring with $A\subseteq B\subseteq\hat{A}$ and $\mathfrak{m}B=\mathfrak{n}$. The conclusion is $\hat{B}\simeq\hat{A}$.
First one shows that $B/\mathfrak{m}^nB\to\hat{A}/\mathfrak{m}^n\hat{A}$ is surjective: ok
Then he says that $B=A+\mathfrak{m}B$: I don't see why: we have clearly $B\supseteq A+\mathfrak{m}B$. For the other, I see that $B\subseteq A+\hat{\mathfrak{m}}=A+\mathfrak{m}\hat{A}$ but we don't have $\mathfrak{m}\hat{A}\subseteq\mathfrak{m}B$. One could say, let $b\in B$, if $b\in\mathfrak{n}=\mathfrak{m}B$ the it's clear, if not $b\in B^*$ then $b=1+(b-1)$ and we have to show $b-1\in\mathfrak{n}$: is there a criterion $1+\beta\in B^*\iff\beta\in\mathfrak{n}$? I don't think because in $\mathbb{Z}_{(p)}$, $1+\frac{1}{3}=\frac{4}{3}$ and $\frac{1}{3}\in\mathbb{Z}_{(p)}^*$ and $\frac{4}{3}\notin p\mathbb{Z}_{(p)}$.
For the containment $B\subset A+\mathfrak{m}B$ can you try taking the quotient $B / (A+\mathfrak{m}B) = B/(A+\mathfrak{n})$ and show the quotient is zero?