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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 $n \in \mathbb{Z}$.

Why $Spec ~R = Proj~R[z]?$

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.

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  • $\begingroup$ Your Proj calculation isn't right. z should be the only variable with nontrivial degree (=1). Proj(k[x][z]) is a P0-bundle over Spec(k[x])=A1, thus isomorphic to A1. $\endgroup$
    – Yosemite Stan
    Commented Jun 17, 2019 at 20:00
  • $\begingroup$ Dear @YosemiteStan, what do you mean by $P0$-bundle? A bundle whose fibers are a point P0? I can't see why is that if were this the case. Thanks in advance. $\endgroup$
    – User43029
    Commented Jun 17, 2019 at 20:07
  • $\begingroup$ Indeed, if I understand it correctly, from the definition $R_i \subset R[z]_i$, then the degree $1$ piece of $k[x]$ (that is, $x$), should be contained in $R[z]_1$ right? $\endgroup$
    – User43029
    Commented Jun 17, 2019 at 20:14

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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 $R[z]_{i+j}$.

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  • $\begingroup$ Thank you! I think I can prove that $Proj~R[z] = Spec~R \otimes_{Spec~k} Proj~k[x] $. My only problem is to see that $x^iz^j \in R[z]_j$, once the grading says that I have $R_i \subset R[z]_i$. Doesn't this mean that $x^i \in R[z]_i$? $\endgroup$
    – User43029
    Commented Jun 18, 2019 at 11:13
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    $\begingroup$ You're welcome, yes there is a grading on $R$ in the case when $R=k[x]$ but when you take Proj R[z] the convention is that you forget the grading that $R$ may or may not have. By definition the grading on $R[z]$ is defined such that all elements in $R\subset R[z]$ are defined to be in degree zero, i.e $R\subset R[z]_0$. The element $z$ is defined to have degree $1$. Thus in the case $R=k[x]$ any element $x^i$ has degree $0$ and $x^i z^j$ has degree $j$. $\endgroup$
    – user130124
    Commented Jun 18, 2019 at 13:11

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