# Trivial Gradation is Gradation

## Theorem

Let $\left({R, +, \circ}\right)$ be a ring.

Let $\left({M, e,\cdot}\right)$ be a monoid.

Let

- $\displaystyle R = \bigoplus_{m \mathop \in M} R_m$

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 $m \neq e$ or $n \neq e$.

After possibly exchanging $m$ and $n$, we may as well assume that $n \neq e$.

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.

$\blacksquare$