Construct the isomorphism from $\operatorname{Proj} A$ to $\operatorname{Proj} A'$

Question

In Gortz, Wedhorn book Algebraic Geometry - Remark 13.7 (p.371)

They say,

Let $A=\bigoplus_{d\ge0}A_d$ be a graded ring. We can "thin out" $A$ and "change $A_0$" without changing the scheme $\operatorname{Proj} A$. More precisely, fix integers $k,\delta\ge1$ and define a new graded ring $A'$ by $A_0'=\mathbb Z$, $A'_d=0$ for $0<d<k$ and $ A'_d=A_{d\delta}$ for $d\ge k$. $\mathfrak{p}\mapsto \mathfrak{p}\cap A'$ defines a bijection $\operatorname{Proj} A\to \operatorname{Proj} A'$. For any homogeneous element $f\in A_+$ we find $f^{k\delta}\in A'$. It is clear that $D_+(f)=D_+(f^{k\delta})$ and it is easy to see that $A_{(f)}=A'_{(f^{k\delta})}$. Thus we have an isomorphism $$\operatorname{Proj} A\to\operatorname{Proj} A'$$

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[Question and try]

I wonder if the map $\mathfrak{p} \mapsto \mathfrak{p} \cap A'$ is actually $\mathfrak{p}= \bigoplus_d \mathfrak{p}_d \mapsto \bigoplus_d(\mathfrak{p}_{d\delta} \cap A_{d\delta}$). Then, in $d=0$ term, $\mathfrak{p}_{0} \cap A_0 = \mathfrak{p}_0 \cap \mathbb{Z}$ ? I think it is nonsense. In fact, I don't understand how to intersect $\mathfrak{p}$ with $ A'$

For the simplicity take $k=2 , \delta=2$. Then, maybe $$A= A_0 \;\oplus \;A_1\;\oplus \;A_2\;\oplus \;A_3\;\oplus \;A_4\;\oplus \;A_5 \;\oplus A_6\;\oplus \;\dots $$ $$A'=\mathbb{Z}\;\oplus \;\;0\;\; \;\oplus A_4\;\oplus \;A_6\;\oplus \;A_8 \;\oplus \;A_{10} \;\oplus A_{12} \;\oplus \;\dots$$

In this situation, how can I construct graded homomorphism from $A$ to $A'$? That is, how can I have an isomorphism from $\operatorname{Proj} A\to\operatorname{Proj} A'$.

Thank you.

Answer 1

One thing you have backwards here is that you don't want a map $A\to A'$, you want a map $A'\to A$ (every nonzero ring has a nonzero map from $\Bbb Z$ because it's an initial object, but not that many rings have a nonzero map to $\Bbb Z$: no field does, for instance, so $A_0$ can't be a field if you want a map $A\to A'$).

You are almost correct with your definition of $p'$: the way we get $p'\subset A'$ is by letting $(p')_d = p_{d\delta}$ for $d>0$, and letting $(p')_0$ be the intersection of $p$ with the image of the canonical map $\Bbb Z\to A_0$ given by sending $1\mapsto 1$.

From here you can just apply the argument from above the "Question and try": if $D(f_i)$ for some collection of $f_i$ homogeneous of positive degree cover $\operatorname{Proj} A$, then $D(f_i^d)$ cover $\operatorname{Proj} A'$ for the same collection of $f_i$, and the coordinate algebras of these affine pieces are the same and have the same gluing data.