Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi : A\rightarrow B$ is a map such that $\phi(\mathfrak{m}) \subset \mathfrak{n}$ and suppose
- $A/\mathfrak{m} \cong B/\mathfrak{n}$;
- $\mathfrak{m} \rightarrow \mathfrak{n/n^2}$ is surjective;
- $B$ is a finitely generated $A$ module.
Show that $\phi$ is surjective.
Here is my approach so far: Consider $0\rightarrow \mathfrak{m} \rightarrow A \rightarrow A/\mathfrak{m}\rightarrow 0$ and the exact sequence for $B$. Then we can establish an commutative diagram of the two exact sequences. (Sorry i do not know how to type that.) Then apply the snake's lemma, the isomorphism in condition 1 yields an isomophism $\mathfrak{n}/\phi(\mathfrak{m})\cong B/\phi(A)$. To see $\phi$ is surjective, it suffices to show $\phi(\mathfrak{m})=\mathfrak n$. Since $\mathfrak{n}$ is the maximal ideal of $B$, if $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$, Nakayama will give the desired result. Now somehow I want to conclude $\phi(\mathfrak{m})+\mathfrak{n}^2 = \mathfrak{n}$ from conditions 2) and 3), which seems to be true but cannot really proceed.