I was wondering if there exist commutative or non-commutative rings for which no non-trivial gradings exist? That is, a ring for which if $R=\bigoplus_{i \in \mathbb{Z} }R_i$ where $R_i$ are abelian subgroups of $R$ and $R_iR_j \subseteq R_{i+j}$, then $R = R_0$ and $R_i = \{0\}$ otherwise.

Initially I believed such a ring would have to be non-commutative, as if it were commutative and we picked some generating set $\{x_i \mid i \in I\}$, then we could declare the $x_i$ to be of degree one. I then realised the degree of some product of the $x_i$ could be expressible in many ways, so the degree of this product might not be well-defined.

But either way, does such a ring with no grading exist? And if not, what about if we restrict to $\mathbb{N}$-gradings?

As I write this I realise that $\mathbb{F}_{p^m}$ must have all elements as degree zero, as if not, and $x$ is in the $n$th graded part, then it is also in the $np$ graded part as $x^{p^m} = x$, but then we have $n(p^m-1) = 0$, so $n=0$. So are there any examples which are not finite fields?

Thank you very much for any help.

How about $\mathbb Z$? It can't even be written as a nontrivial direct sum of subgroups. (Proof: if $a \in \mathbb Z$ is in one direct summand and $b$ is in another, then $ab$ must be in both.)