Trivial Gradation is Gradation
Theorem
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {M, e, \cdot}$ be a monoid.
Let
- $\ds R = \bigoplus_{m \mathop \in M} R_m$
be the trivial $M$-gradation on $R$.
This is a gradation on $R$.
Proof
We are required to show that:
- $\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$
First suppose that $m = n = e$ are both the identity.
In this case, $R_m = R_n = R$.
Since by definition, $R$ is closed under $\circ$, it follows that
- $\forall x \in R, y \in R: x \circ y \in R$
as required.
Now suppose that either $m \ne e$ or $n \ne e$.
After possibly exchanging $m$ and $n$, we may as well assume that $n \ne e$.
In particular, $R_n = \mathbf 0$ is the zero ring.
So if $y \in R_n$, then $y = 0$.
Therefore, for every $x \in R_m$, by Ring Product with Zero, we must have
- $x \circ y = x \circ 0 = 0$
Since $R_{m \cdot n}$ is an abelian group it must by definition contain $0$.
Therefore $x \circ y \in R_{m \cdot n}$ as required.
$\blacksquare$