Proj of graded rings

Question

my question actually concerns an exercise II5.13 in Hartshorne. You have a graded ring $S=\oplus S_n$ with $n\ge0$ generated as $S_o$-Algebra by $S_1$ and you set $S^{(d)}=\oplus S_{dn}$ for a $d>0$.

Why is then $Proj(S) \simeq Proj(S^{(d)})$ ?

Just give me some hints, that would be nice!

Answer 1

Let $A$ be a $\mathbb{N}$-graded ring: $A = \bigoplus_{i \geq 0} A_i$. Fix $d > 0$ and consider $A^{(d)} = B = \bigoplus_{i \geq 0} A_{id} \subseteq A$ as sets, but with different graduations: $A^{(d)}_i = A_{di}$ and

• $B_i = 0$ if $d$ does not divide $i$;
• $B_i = A_i = A^{(d)}_{i/d}$ if $d$ divides $i$.

1) Prove that $Proj A^{(d)} \simeq Proj B$.

2) Observe that $B$ is a graded subring of $A$ and consider the morphism $\phi$ induced by the inclusion $B \hookrightarrow A$. Prove that $\phi$ is an isomorphism between $Proj \ B$ and $Proj \ A$. (Hint: what is the domain of $\phi$? How does $\phi$ act on principal affine open subsets?)

Answer 2

This follows from Proposition 2.3.38 in Liu's book Algebraic geometry and arithmetic curves. I don't know the reference for this theorem in Hartshorne's book, but it shouldn't be too hard to figure out.

Answer 3

The canonical inclusion $S^{(d)}\to S$ gives you a morphism $\mathrm{Proj}(S^{(d)})\to\mathrm{Proj}(S)$. We can check on stalks that it is an isomorphism; so let $\mathfrak{p}$ be a homogeneous prime ideal of $S$ which is not irrelevant and $\mathfrak{q}:=S^{(d)}\cap\mathfrak{p}$. Then, $(S_{\mathfrak{p}})_0\cong(S^{(d)}_{\mathfrak{q}})_0$ by choosing any element $g\in S_d\setminus\mathfrak{p}_d$ and expanding fractions, i.e. $a/f\mapsto ag/fg$.