In the following, let "ring" be a synonym for "commutative ring with identity". In the book on Commutative Algebra by Atiyah and MacDonald, I read:

Let $A$ be a local ring, $\mathfrak{m}$ its maximal ideal, $k = A / \mathfrak{m}$ its residue field. Let $M$ be a finitely generated $A$-module. $M / \mathfrak{m}M$ is annihilated by $\mathfrak{m}$, hence is naturally an $A/\mathfrak{m}$-module, i.e. a $k$-vector space, and as such finite-dimensional.

Proposition 2.8: Let $x_i$ $(1 \leq i \leq n)$ be elements of $M$ whose images in $M / \mathfrak{m}M$ form a basis of this vector space. Then the $x_i$ generate $M$.

In the proof, the fact that $A$ is local is not used explicitely. I suspect that this holds for any non-trivial ring $A$ and some maximal ideal $\mathfrak{m}$ of $A$. However, I might overlook something.

Question: Does Proposition 2.8 hold for any non-trivial ring $A$ with maximal ideal $\mathfrak{m}$? If no, why is it important for $A$ to be local?

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    $\begingroup$ The implication $N+\mathfrak mM=M\implies N=M$ requires $\mathfrak m$ to be contained in the Jacobson radical of $A$. $\endgroup$ Apr 22, 2013 at 3:30
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    $\begingroup$ Dear Corsin, Did you consider the case $A =\mathbb Z$, $\mathfrak m = 2\mathbb Z$, and $M = \mathbb Z/3\mathbb Z$? Then $M/\mathfrak m M$ is generated by zero elements (i.e. is the trivial module), but this is not true of $M$ itself. Regards, $\endgroup$
    – Matt E
    Apr 22, 2013 at 3:34

1 Answer 1


If $A$ is a ring (commutative with identity), $I$ an ideal of $A$, and $M$ a finitely generated $A$-module, then $M/IM$ is a finitely generated $(A/I)$-module. In fact, if $m_1,\ldots,m_r$ generate $M$ as an $A$-module, then there images in $M/IM$ generate it as an $(A/I)$-module. If $I$ is contained in the Jacobson radical of $A$ (meaning $I$ is contained in every maximal ideal of $A$) and $m_1+IM,\ldots,m_r+IM$ generate $M/IM$ as an $(A/I)$-module, then $m_1,\ldots,m_r$ generate $M$ as an $A$-module (assuming $M$ is finitely generated as an $A$-module). This is an application of a general form of Nakayama's lemma, which is also used to prove the special case where $A$ is local and $I=\mathfrak{m}$ is the unique maximal ideal (for $A$ local, its Jacobson radical is the maximal ideal).

Namely, let $N$ be the $A$-submodule of $M$ generated by $m_1,\ldots,m_r$. We then have $I(M/N)=M/N$. Indeed, given $m\in M$, by assumption, we have $m+IM=\sum_{i=1}^ra_im_i$ for some $a_i\in A$. Thus $m-\sum_{i=1}^ra_im_i\in IM$. This gives the non-trivial inclusion $M/N\subseteq I(M/N)$. By Nakayama, there is $a\in A$ with $a\equiv 1\pmod{I}$ such that $a(M/N)=0$. But because $I$ is contained in the Jacobson radical of $A$, $a$ must be a unit, and therefore $M=N$.

  • $\begingroup$ The special case of Nakayama can be proven directly. $\endgroup$ Apr 22, 2013 at 7:42

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