# $\mathbb{Z}$-graded ring with no homogeneous prime ideals is isomorphic to Laurent polynomial ring

If $R$ is a $\mathbb{Z}$-graded ring with no nonzero homogeneous prime ideals, I want to show $R$ is isomorphic to $R_0[x,x^{-1}]$ unless $R=R_0$.

Now if I take $0\neq x \in R_1$, $x$ has an inverse (since if not, $(x)$ is contained in a homogeneous prime), but why is $R_0[x,x^{-1}]$ everything?

If $x\in R_1$ is a unit, then notice that for any $n$ and any $y\in R_n$, $x^{-n}y\in R_0$. Writing $r=x^{-n}y$, we then have $y=rx^n\in R_0[x,x^{-1}]$.
(Note, though, that this doesn't finish the proof. First of all, this only shows that the canonical homomorphism $R_0[x,x^{-1}]\to R$ is surjective; you also have to show it is injective. Second, what if, say, $R_1=0$ but $R_2\neq 0$? As a hint to handle the general case, you'll want to take a nonzero $x\in R_n$ for the least positive $n$ such that $R_n\neq 0$, and then you'll need to show $R_m=0$ whenever $m$ is not a multiple of $n$.)