Quick question about Graded Rings

Question

Recently I read about graded rings and I read old papers but I noticed something all these papers define the graded ring but there is no proves (all rings are a group graded ring and satisfy the condition $R_g.R_h \subseteq R_g+R_h$). So,

Is there any ring that's not a $G$-graded ring?

If the answer is no, how can I proof that the inclusion condition is satisfied for any ring and group $G$ (I mean here the non-trivial graded ring)?

Maybe my question would be trivial, but I need a clear vision.

No, Lobo: the standard definition requires only that $G$ is a semigroup (or monoid if $R$ has identity) — Eric Stucky, May 08 '17 at 12:24

score 1 · Accepted Answer · answered May 08 '17 at 13:04

1

We claim that $\Bbb Z$ is not a $G$-graded ring for any nontrivial monoid $G$.

Suppose it is; then in particular $\Bbb Z=\bigoplus A_g$ for some subgroups $A_g$ of $\Bbb Z$ with $A_1\neq\Bbb Z$. But then the injection $A_1\to\Bbb Z$ is given by multiplication by some number $m$, which implies that $A_*=\bigoplus_{g\neq 1} A_g$ has a nonzero torsion element (since, e.g. $0\to A_0\to\Bbb Z\to A_*\to 0$ is a split exact sequence). But since $\Bbb Z$ has no torsion, we have a contradiction.

The proof goes through for semigroups, of course, since there was nothing essential used about the identity element here.

answered May 08 '17 at 13:04

Eric Stucky

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Good, but G here is monoid what about if G is a group ? – raindrops May 11 '17 at 08:45
I am sure you can work this out for yourself. Try to copy the argument. – Eric Stucky May 11 '17 at 22:46
@raindrops : any group is a monoid , so this answer works equally well for groups too ... or am I missing something ? – user Jun 25 '17 at 07:53
(yes, @users, that is what I was getting at.) – Eric Stucky Jun 25 '17 at 15:24

Quick question about Graded Rings

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