I am reading Thaddeus' paper on GIT and flips (<https://arxiv.org/pdf/alg-geom/9405004.pdf>), and I am confused with a claim in the begining. 


Let $R$ be a finitely generated integral algebra over an algebraically closed field $k$, and $X = Spec~R$. Choose a $\mathbb{Z}$-grading on $R$. (It is possible to find such grading because it is possible to define a $k^\*$-action on $X$). Let $z$ be an indeterminate, and define a $\mathbb{Z}-$ grading on $R[z]$ by $R\_i \subset R[z]\_i$, and letting $z \in R[z]\_{-n} $ by some $n \in \mathbb{Z}$.



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> Why $Spec ~R = Proj~R[z]?$
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Clearly, this has something to do we the grading on $R[z]$, because, if $R = k[x]$, then with the usual grading on $k[x][z]$ we would have $\mathbb{A}^1 = Spec~R = Proj~R[z] = Proj~k[x][z] = \mathbb{P}^1$, a contradiction. I am failing to see in this example, how the grading really works. 


Any help would be appreciated. 



Relation between the Spec and the Proj of a ring
Clearly $\text{Proj}k[z]\simeq \text{Spec}k$ (projective $0$-space is a point) and thus $\text{Proj}R[z]\simeq \text{Spec}R\times\_{\text{Spec}k}\text{Proj}k[z]\simeq \text{Spec}R\times\_{\text{Spec}k}\text{Spec}k\simeq \text{Spec}R$. The grading on $R[z]$ when $R=k[x]$ is not the grading on $k[x,z]$ by degree, $x^iz^j\in R[z]\_j$ and not $R[z]\_{i+j}$.


