My friend and I were working through old preliminary exams. This problem is the fifth one on this list, but we were unable to come up with a solution.
Let $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ be local Noetherian rings. Suppose that $\phi:A\to B$ is a map such that $\phi(\mathfrak{m})\subset\mathfrak{n}$, and suppose that
- $A/\mathfrak{m}\to B/\mathfrak{n}$ is an isomorphism.
- $\mathfrak{m}\to\mathfrak{n}/\mathfrak{n}^2$ is surjective.
- $B$ is finitely generated as an $A$-module.
Prove that $\phi$ is surjective.
Does anyone have an idea on how to proceed, or possibly a solution to this?
This follows from Nakayama's lemma. By (1) it is enough to show that $\mathfrak n\subset\operatorname{im}(\phi)$:
First off, notice that by (3) and the hypothesis that $A$ is Noetherian, $\mathfrak n$ is a finitely generated $A$-module; a fortiori, it is a finitely generated ideal of $B$. Pick any finite set $S\subset\mathfrak n$ such that its projection $\bar S$ is a basis of $\mathfrak n/\mathfrak n^2$ over $B/\mathfrak n$. View it as a basis over $A/\mathfrak m$ using (1). Then, by Nakayama's lemma, $S$ is a generating set for $\mathfrak n$ over $A$. By (2), we can pick $x_1,\dots,x_r\in A$ such that $\overline{\phi(x_1)},\dots,\overline{\phi(x_r)}$ generates $\mathfrak n/\mathfrak n^2$. Setting $S=\{\phi(x_1),\dots,\phi(x_r)\}$, we see that $\phi$ surjects onto $\mathfrak n$.