Let $R$ be a $\mathbb Z$-graded ring and let $M$ be graded $R$-module. Show that $x \in R$ is homogeneous element of nonzero degree then $1-x$ is a nonzero divisor on $M$.
I showed that if an element $m \in M$ is homogeneous and $(1-x)m=0$, then $m=0$.
Let $m=m_a +\cdots + m_{a+n} \in M$ where $n \ge 1$ and $\deg m_{a+i} = a+i$.
If the degree of $x$ is strictly greater than $n$, then we have all $m_a, \dots, m_{a+n} =0$.
However, when $0 < \deg x \le n$, I am stuck.
How can we show that $m=0$?
Suppose there is some $m$ such that $(1-x)m = 0$. Denote by $m_i$ the $i$th homogeneous component of $m$. Let the "bottom degree" of $m$ be the smallest $k$ such that $m_k \neq 0$.
First assume that the degree of $x$ is positive. Then just consider the bottom degree term of $(1-x)m$: it must be $m_k$ since $x$ is homogeneous of degree $>0$. Thus $(1-x)m =0$ would require $m_k = 0$, contradicting our assumption that it is nonzero.
In the case that the degree of $x$ is negative, consider instead the top degree term.