Let $A\rightarrow R$ be a homomorphism of commutative rings with identity. Define the $\mathbb{Z}^{\geq 0}$ graded ring $$S=A\oplus R \oplus R \oplus R \oplus...$$ with multiplication defined in the only sensible way.
It is claimed that $$Proj ~S\simeq Spec ~ R.$$How to prove that?
For $r\in R$ let me write $[r]_k$ for the element of the k'th-degree summand of $S$. Since $$Proj ~ S = \cup_{f\in R}D([f]_1)$$It seemed to me that the logical thing to do was define local maps (for each $f\in R$) by $$D([f]_1)=Spec ~(S_{[f]_1})_0\rightarrow Spec ~R$$ $$R~\rightarrow (S_{[f]_1})_0$$ by $$r\mapsto \frac{[rf]_1}{[f]_1}.$$ These indeed seem to patch together and to factor through local isomorphisms $$R_f~\rightarrow (S_{[f]_1})_0. $$ The problem is that then we have in particular $$R =R_1 \simeq (S_{[1]_1})_0$$ but we needn't have $$D([1]_1)= Proj~S.$$ So that when $D([1]_1)\neq Proj~S$ the underlying map of points for the proposed isomorphism will not be bijective. What am I doing wrong?
The key idea here is that $S$ looks a lot like $R[x]$ and it's obvious that $\operatorname{Proj} R[x] = \operatorname{Spec} R$. So if we can show that $\operatorname{Proj} R[x] \cong \operatorname{Proj} S$, we'll be done.
The idea here is that given a homogeneous non-irrelevant prime $\mathfrak{p}\subset R[x]$, we know what it should correspond to in $S$: $\mathfrak{p}\cap S$, where we identify $x\in R[x]$ with $[1]_1$ using your notation. Why does this produce a bijection on the required ideals? The answer is that we can determine what should happen in degree $0$ from the structure of the homogeneous ideal in larger degrees: for any $f$ a homogeneous element of positive degree not in $\mathfrak{p}$ (which exists since $\mathfrak{p}$ is non-irrelevant), we can recover $\mathfrak{p}_0$, the zeroth-graded piece of $\mathfrak{p}$ as the set $\{x\in R_0 \mid xf^a\in \mathfrak{p}_{a\deg f} \forall a > 0\}$. So we've shown that $\operatorname{Proj} R[x]$ and $\operatorname{Proj} S$ have the same points.
To show that they have the same scheme structure, it suffices to show that the bijection plays nicely with the formation of the open sets $D(f)$ for $f$ homogeneous of positive degree and that the ring of sections on $D(f)$ is the same on both sides. But the first part is more or less immediate from the definitions: $D(f)$ is the set of homogeneous non-irrelevant prime ideals not containing $f$ (of positive degree), so everything is phrased in terms of what happens in positive degree, and the bijection described above acts as the identity on the positive-degree parts. To show that the ring of sections we assign to each $D(f)$ is the same, note that any element of the ring of sections can be represented as a quotient of elements of the same positive degree by multiplying by $\frac{f}{f}$. So this also depends only on the positive degree portions of our ring, and we're done.
Remark: similar tactics can be used to show that for a graded ring $R$ and a fixed positive integer $n$, the ring $R'$ which has $d^{th}$ graded piece $R_d$ if $d$ is a multiple of $n$ and $0$ otherwise also satisfies $\operatorname{Proj} R = \operatorname{Proj} R'$.