Nakayama's lemma

Remarks

Proof idea: let $m_1, \ldots, m_n$ generate $M$ with $n$ minimal, and suppose $n \ge 1$. From $IM = M$, write $m_n = a_1 m_1 + \cdots + a_n m_n$ with $a_i \in I$. Then $(1 - a_n) m_n$ lies in the submodule generated by $m_1, \ldots, m_{n-1}$. Since $a_n \in J(R)$, the element $1 - a_n$ is a unit, so $m_n$ is not needed, contradicting minimality. Another proof uses the determinant trick to find $a \in I$ with $(1 + a) M = 0$.

Consequence: if $N \subseteq M$ is a submodule and $M = N + IM$, then $M = N$; apply the lemma to $M/N$. For a local ring $(R, \mathfrak m)$ with $I = \mathfrak m$, this says that elements of $M$ generate $M$ exactly when their images span the vector space $M/\mathfrak m M$ over $R/\mathfrak m$.

Both hypotheses are needed. Without finite generation: over $R = \Z_{(p)}$ with $I = p\Z_{(p)} = J(R)$, the module $M = \Q$ satisfies $pM = M$ but $M \ne 0$. Without $I \subseteq J(R)$: over $R = \Z$ with $I = 2\Z$, the module $M = \Z/3\Z$ satisfies $2M = M$, since $2$ is invertible modulo $3$, but $M \ne 0$.

Going deeper

For a student

The lemma is the main tool for passing from the residue field back to the local ring. It gives the equality between the minimal number of generators of a finitely generated module $M$ over a local ring and $\dim_k M/\mathfrak m M$. With $M = \mathfrak m$, this identifies the embedding dimension with the minimal number of generators of $\mathfrak m$. It also shows that finitely generated projective modules over local rings are free, and it enters the proof of the Krull intersection theorem.