In a $\mathbb{Z}$-graded ring with unity, $1$ is homogeneous of degree zero

Question

Suppose I have a $\mathbb{Z}$-graded ring (commutative) $A$ with unity. I am sure that the unity $1 \in A$ has degree $0$. I was wondering how could one show that?

(I am guessing we don't have to take it as an axiom...)

Answer 1

Write $1=a_p+\cdots+a_q$ with $p\le q$, $a_p\ne 0$ and $a_q\ne 0$. It is easily seen that we can't have $q<0$ or $p>0$, otherwise multiply $1=a_p+\cdots+a_q$ with an arbitrary homogeneous element and get that every homogeneous element of $A$ is $0$, a contradiction.

Then $q\ge 0$ and $p\le 0$, and assume that at least one of the two inequalities is strict. Say $q>0$. From $1=a_p+\cdots+a_q$ we get $x=xa_p+\cdots+xa_q$ for every homogeneous $x\in A$, hence $xa_q=0$ for every homogeneous $x\in A$. In particular, $a_ia_q=0$ for all $i=p,\dots,q$, hence $1\cdot a_q=0$, a contradiction.

Conclusion: $p=q=0$, and therefore $1\in A_0$.

Answer 2

If $a\in A_n$ is of degree $n$, what is the degree of $a^2$?