how to prove a graded version of Nakayama's lemma There is a simple version of graded Nakayama lemma in Wikipedia: > If $R$ is a positive graded ring, let $M$ be a graded $R$ module such that $M_i=0$ for $i$ sufficiently negative (such as $M$ is finite generated). If $R_{+}M=M$, then $M=0$. I am confused one version of graded Nakayama lemma in _Cohen-Macaulay Rings_ , Exercise 1.5.24. > Let $R$ be a ring with a unique graded maximal ideal $m$ (i.e. $m$ is a graded ideal and every proper graded ideal is contained in $m$). Let $M$ be a finite generated graded module such that $mM=M$. Then $M=0$. How to prove this? Thank you in advance.

Since $M$ is finite generated graded module,we can select homogeneous element $x_1,x_2,\cdots x_n$ such that they generate $M$.By $mM=M$,there exists homogeneous elements $a_{ij},1\leq i,j\leq n\in m$,where $deg(a_{ij})=deg(x_i)-deg(x_j)$ such that $a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n=x_i,i=1,2,\cdots,n$.Hence $-a_{i1}x_1-a_{i2}x_2+\cdots +(1-a_{ii})x_i+\cdots-a_{in}x_n=x_i,i=1,2,\cdots,n$.Through multiply the adjoint matrix,we get $(detB)M=0$,where $B=(b_{ij})$ is the matrix of coefficients.remark that $detB=\sum_{\sigma\in S^n}b_{1\sigma(1)}b_{2\sigma(2)}\cdots b_{n\sigma(n)}$,$deg(b_{1\sigma(1)}b_{2\sigma(2)}\cdots b_{n\sigma(n)})=deg(x_1)-deg(x_{\sigma(1)})+deg(x_2)-deg(x_{\sigma(2)})+\cdots+deg(x_n)-deg(x_{\sigma(n)})=0$.

it is clear that $detB=1+a$ where $a\in m$ is homogeneous of degree 0.since $m$ is maximal graded ideal,we know $1+a$ is a unit.