Let $G$ be a monoid ; $R$ be a unital commutative $G$-graded ring such that for every $g \in G$ , $x_gy\ne 0 , \forall x_g \in R_g \setminus \{0\} , \forall y \in R \setminus \{0\}$ . Then is it true that $R$ is an integral domain ?
I am only able to show that if $G$ is a cancellative monoid and $x , y \in R$ be such that $xy=0$ then , either $y \notin \cup_{g \in G} R_g$ or else $y=0$ .
If the claim is not true in general , then is there any extra condition on $G$ which will make the claim true ? Like $G$ being a finite group or a torsion free abelian group ?
Please help . Thanks in advance
1) No when $G$ is a finite group. A very simple counterexample (with $G$ the group on 2 elements, written $\{0,1\}$ modulo 2):
over a field, the group algebra of $G$. So it has the basis $(x_0,x_1)$ with commutative law $$x_0x_0=x_0,\quad x_0x_1=x_1,\quad x_1x_1=x_0.$$
The grading is the obvious one ($x_i$ of degree $i$ modulo 2). Zero divisors are scalar multiples of $x_0+x_1$ and $x_0-x_1$; these are not homogeneous.
2) Yes when $G$ is a torsion-free abelian group. Indeed, endow $G$ with a total ordering (stable under translation) and consider a $G$-graded ring (ring in the weakest sense: abelian group with bilinear law). Given $x,y\neq 0$ with $xy=0$. Let $F_x$ be the support of $x$ (the set of $g$ such that $x_g\neq 0$); it is finite. Let $g=\max(F_x)$, $h=\max(F_y)$. Then expanding $xy=0$, we see that $x_gy_h\neq 0$, with $x_g,y_h\neq 0$.
The existence of such a total ordering on $G$ is not hard to see, but here we just need it on the subgroup of $G$ generated by $F_x\cup F_y$, which is isomorphic to $\mathbf{Z}^k$ for some $k$ and there we can take a lexicographic order, or the order induced by some group embedding into $\mathbf{R}$.