Hartshorne Exercise II.5.13.
Let $S$ be a graded ring, generated by $S_1$ as an $S_0$ -algebra. For any integer $d>0$, let $S^{(d)}$ be the graded ring $\bigoplus_{n\ge 0} S_{n}^{(d)}$ where $S_n^{(d)}=S_{nd}$. Let $X = \op{Proj}S$. Show that $\op{Proj}S^{(d)}=X$, and that the sheaf $\OO(1)$ on $\op{Proj}S^{(d)}$ corresponds via this isomorphism to $\OO_X(d)$.
This construction is related to the $d$-uple embedding (I, Ex. 2.12) in the following way. If $x_0,…,x_r$ is a set of generators for $S_1$, corresponding to an embedding $X\ra\PP{A}{r}$, then the set of monomials of degree $d$ in the $x_i$ is a set of generators for $S_1^{(d)}=S_d$. These define a projective embedding of $\op{Proj}S^{(d)}$ which is none other than the image of $X$ under the $d$-uple embedding of $\PP{A}{r}$.
We construct explicitly an isomorphism $\op{Proj}(S)\longrightarrow\op{Proj}(S^{(d)})$. Indeed, consider the Veronese Subring,
$$\iota:S^{(d)}\xrightarrow{\subseteq}S.$$
It is clear that this is an integral extension of rings, and moreover that homogeneous primes of $S$ that don’t contain $S_+$ pull back to homogeneous primes of $S^{(d)}$ that don’t contain $S^{(d)}_+$. Moreover, if two homogeneous primes $\p,\p’\subset S$ that don’t contain all of $S_+$ pullback to the same prime $\q\subset S^{(d)}$, then I claim that $\p=\p’$. So, we have
$$\bigoplus_{n\in \N}(S_{nd}\cap\p) =\bigoplus_{n\in \N}(S_{nd}\cap\p’),$$
but suppose for a contradiction that $\p\ne \p’$. WLOG, $f\in \p-\p ‘$. Let $m\ge 1$ be the smallest integer such that $S_m\cap \p\ne S_m\cap\p’$. Note that $d\not|m$. Let $nd>m$ be the smallest multiple of $d$ that is larger than $m$ such that: $S_{nd}\cap \p\ne\varnothing$, $S_{nd-m}\ne\varnothing$, and $S_{nd-m}\ne S_{nd-m}\cap \p’$. Such an integer $nd$ must exist by taking $n>>0$ (for example, take $n=m$). Let $f\in \p-\p’$ be degree $m$ and let $g\in S_{nd-m}-\p’$. Then $gf\in S_{nd}\cap\p=S_{nd}\cap\p’$, which implies that actually $f\in\p’$, and we reach our contradiction. This shows that we have a continuous bijection of schemes,
$$f:\op{Proj}S\longrightarrow\op{Proj}S^{(d)}.$$
To show this is a closed map, take any homogeneous ideal $I\subset S$; we certainly have $f(V(I))\subseteq V(I\cap S^{(d)})$. Suppose that $\q\supset I\cap S^{(d)}$ is a homogeneous prime in $S^{(d)}$. There is some homogeneous prime $\p\subset S$ lying over $\q$ by integrality, and we must show that $I\subset \p$. We write
$$\bigoplus_{n\in \N}(S_{nd}\cap I)\subset \bigoplus_{n\in \N}(S_{nd}\cap \p).$$
Proceeding by contradiction and following an identical argument to that above gives the desired result. We therefore conclude that $f$ is an isomorphism of schemes. We should breifly note that this indeed an isomorphism of schemes, not just topological spaces, because the morphism $\iota$ on structure sheaves is a local isomorphism at the stalk of every homogeneous prime:
$$\iota:(S^{(d)})_{(\iota^{-1}(\p))}\xrightarrow{\cong} S_{(\p)}.$$
To see this, we just need to check surjectivity, which holds because we can always do something like $\frac{xy}{y^2}=\frac xy$. More generally, if $\frac{g}{h}\in S_\p$ with $h\notin\p$ and $h,g$ are homogeneous of the same degree $m\ge 1$ not a multiple of $d$, then we see that
$$\frac{gh^{d-1}}{h^d}\longmapsto \frac{g}{h}.$$
Under this isomorphism, $\OO(1)=\tilde S_1$ on $\op{Proj}S^{(d)}$ pulls back to $\OO(d)=\tilde S_d$ on $\op{Proj}S$. The $d$-uple embedding of $\op{Proj}S\ra \PP{A}{N}$ factors as $\op{Proj}S\ra \op{Proj}S^{(d)}\subset \PP{A}{N}$, under which $\OO_{\PP{A}{N}}(1)$ first restricts to $\OO_{\op{Proj}S^{(d)}}(1)$ (from the surjection $A[y_0,…,y_N]$ that sends $y_i\mapsto M_i(x_0,…,x_r)$), and then pulls back to $\OO_{\op{Proj}S}(d)$.
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