# equivalent formulation of Nakayama’s lemma

The following is equivalent^{} to Nakayama’s lemma.

Let $A$ be a ring, $M$ be a finitely-generated $A$-module, $N$ a submodule of $M$, and $\U0001d51e$ an ideal of $A$ contained in its Jacobson radical^{}. Then $M=\U0001d51eM+N\Rightarrow M=N$.

Clearly this statement implies Nakayama’s Lemma, by setting $N$ to $0$. To see that it follows from Nakayama’s Lemma, note first that by the second isomorphism theorem for modules,

$$\frac{\U0001d51eM+N}{N}=\frac{\U0001d51eM}{\U0001d51eM\cap N}$$ |

and the obvious map

$$\U0001d51eM\to \U0001d51e\frac{M}{N}:am\mapsto a(m+N)$$ |

is surjective^{}; the kernel is clearly $\U0001d51eM\cap N$. Thus

$$\frac{\U0001d51eM+N}{N}\cong \U0001d51e\frac{M}{N}$$ |

So from $M=\U0001d51eM+N$ we get $M/N=\U0001d51e(M/N)$. Since $\U0001d51e$ is contained in the Jacobson radical of $M$, it is contained in the Jacobson radical of $M/N$, so by Nakayama, $M/N=0$, i.e. $M=N$.

Title | equivalent formulation of Nakayama’s lemma |
---|---|

Canonical name | EquivalentFormulationOfNakayamasLemma |

Date of creation | 2013-03-22 19:11:47 |

Last modified on | 2013-03-22 19:11:47 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 4 |

Author | rm50 (10146) |

Entry type | Theorem |

Classification | msc 13C99 |