Zero direct summand

Question

I want a suggestion for the following:

Let $M$ be a finitely generated $R$-module, where $R$ is a commutative Noetherian local ring with $1_R$, and $N$ a direct summand of $M$ such that $N⊆mM$, where $m$ is the maximal ideal of $R$. Then $N=0$.

I guess that the lemma of Nakayama works. Thanks a lot!

Answer 1

As your ring is local, this means that $m$ (the maximal ideal) coincides with the Jacobson radical. Furthermore, you can easily verify that, being $N$ a direct summand, then $mN=mM\cap N=N$ (where the second equality is one of your assumptions). Thus, $N$ is a finitely generated (as it is a direct summand of a f.g. module), such that $J(R)N=N$, by the Nakayama Lemma, $N=0$.

If you want an explicit argument for $IN=IM\cap N$ ($I$ any ideal and $N$ a direct summand of $N$):

-- the inclusion $IM\cap N\supseteq IN$ is trivial;

-- the inclusion $IN\supseteq IM\cap N$ can be proved as follows: take $x=im\in IM\cap N$ with $i\in I$ and $m\in M$. Since $N$ is a direct summand, there exists a complement $N'$ such that $M=N'\oplus N$. Let us write $m$ as $m=n+n'$ with $n\in N$ and $n'\in N'$, so $x=im=in+in'\in N$, implies that $in'=0$ and so $x=im=in\in IN$, as desired.

Answer 2

Let us write $M=N\oplus P$, and observe that $N\subset mM$ implies that $N\subset mN$, which implies, by Nakayama's Lemma, that $N=0$.