Nakayama corollary: generators mod the maximal ideal lift

Let $(R,\mathfrak m)$ be a local ring with maximal ideal $\mathfrak m$ (as in maximal-ideal characterization), and let $M$ be a finitely generated $R$-module. Let $k=R/\mathfrak m$ be the residue field, so $M/\mathfrak m M$ is naturally a $k$-vector space.

1. Generators lift. If $m_1,\dots,m_r \in M$ map to elements whose images generate $M/\mathfrak m M$ as a $k$-vector space, then $m_1,\dots,m_r$ generate $M$ as an $R$-module.

1. Vanishing test. If $\mathfrak m M = M$, then $M=0$. Equivalently, $M=0$ if and only if $M/\mathfrak m M=0$.

1. Minimal number of generators. The minimal number of $R$-module generators of $M$ equals $\dim_k(M/\mathfrak m M)$. In particular, $M$ is cyclic if and only if $\dim_k(M/\mathfrak m M)=1$.

1. Free modules over a DVR-like local ring. Let $R=k[[t]]$ with $\mathfrak m=(t)$, and take $M=R^n$. Then $M/\mathfrak m M \cong k^n$ has $k$-dimension $n$, so $R^n$ needs (and has) exactly $n$ generators.

1. A cyclic module detected mod $\mathfrak m$. With $R=k[[t]]$ and $\mathfrak m=(t)$, let $M=R/(t^2)$. Then $M/\mathfrak m M \cong k$, so $\dim_k(M/\mathfrak m M)=1$, and the corollary says $M$ is cyclic (generated by the class of $1$).

1. Localizing the integers. Let $R=\mathbb Z_{(p)}$ with maximal ideal $\mathfrak m=(p)$, and let $M=(p^2)$ viewed as an $R$-module (an ideal). Then $M/\mathfrak m M = (p^2)/(p^3)$ is $1$-dimensional over $R/\mathfrak m \cong \mathbb F_p$, so $M$ is generated by a single element (namely $p^2$).