Let $A$ be a ring, $\Delta$ a commutative monoid whose elements are all cancellable and $(A_\lambda)_{\lambda\in\Delta}$ a graduation of $A$ compatible with the ring structure of $A$. I want to show that $1\in A_0$. Clearly there exists a unique $a\in\bigoplus_{\lambda\in\Delta}A_\lambda$ such that $1=\sum_{\lambda\in\Delta}a_\lambda$.
Let $\mu\in\Delta$ and $x\in A_\mu$. Then $x=x1=\sum_{\lambda\in\Delta}xa_\lambda$. For each $\lambda\in\Delta$, we have $xa_\lambda\in A_\mu A_\lambda\subset A_{\lambda+\mu}$. The author claims that by "comparing the components of degree $\mu$, $x=xa_0$". I am don't understand what this means.
Let $\lambda\in\Delta-\{0\}$. We know already that $xa_\lambda\in A_{\mu+\lambda}$. If I could show that $xa_\lambda$ is also in $A_\mu$, then if $xa_\lambda\ne0$, we must have $\lambda+\mu=\mu$, i.e. $\lambda=0$ contrary to assumption. But how can I show that $xa_\lambda\in A_\mu$?
The "component of degree $\mu$" of $x$ is just $p(x)$ where $p:A\to A_\mu$ is just the projection map associated to the direct sum decomposition $A=\bigoplus A_\lambda$. Since $x\in A_\mu$, $p(x)=x$. But we also have $x=\sum a_\lambda x$ where each term of this sum is homogeneous, so $p(x)$ is the sum of all the terms which have degree $\mu$. That is, it is the sum of all the terms $a_\lambda x$ such that $\lambda+\mu=\mu$. The only such $\lambda$ is $0$, so $p(x)=a_0x$. Thus $a_0x=x$.