I am confused by an exercise from Tom Marley which is:
Let $R$ be an arbitrary $\mathbb Z$-graded domain:
$1)$ Prove that all units in $R$ are homogeneous.
$2)$ By using $1$, if $R$ is a field, then $R_0=R$ and $R_n=0$ for all $n\neq 0$.
If we remove the domain condition, I know $1$ is not true, but I cannot find the counter example.
Any comments and guidance would be highly appreciated.
If $a$ is a nonzero element in your ring, you write $a=a_n+a_{n+1}+\cdots+a_m$ for some homogeneous $a_i$, with $n\leq m$ and $a_n\neq0$ and $a_m\neq 0$. Call the number $m-n$ the width of $a$ and write it $w(a)$.
Now show that if $a$ and $b$ are nonzero, then $w(ab)=w(a)+w(b)$.
Show using that that a unit has width $0$, so it is homogeneous.
For the second part, if $R$ is a field, then $R$ is a domain and any non-zero element of $R$ is an unit. By the answer given by Mariano above, we know that all the units has degree zero. Thus applying the first part, we write $R=R_0$.
For the counter example of part $1$ where the ring is not a domain, consider the graded ring $\dfrac{R[X]}{(X^n)}$ with standard grading and the element $1+x\in\dfrac{R[X]}{(X^n)}$, where $x=X+(X^n)\in\dfrac{R[X]}{(X^n)}$. Note that as $x$ is nilpotent, $1+x$ is an unit. But $1+x$ is not a homogeneous element.
Let $x,y\in R$ with $xy=1$ and $x=\sum x_i, y=\sum y_j$. We know that $1\in R_0$. So, $x_iy_j=0$ for all $i,j$ with $i+j\neq 0$. Let $x$ be not homogeneous, and has two non zero components $x_{i_a}, x_{i_b}$, where $i_a\neq i_b$. It is clear that $j+i_a\neq 0$ or, $j+i_b\neq 0$ f0r all $j$. Hence, $y_j=0$ for all $j$ and so $y=0$, which is a contradiction.
