# Solutions to Hartshorne's Algebraic Geometry/Projective Morphisms

[Hartshorne 5.14]

Let $X$ be a subscheme of projective space $P^{r}_{A}$, where $A$ is a ring. We define the \textbf{homogeneous coordinate rings} $S$ of $X$ for the given embedding to be $A[x_{0},\cdots,x_{r}]/I$, where $I$ is the ideal $\Gamma_{\ast}(\mathscr{I}_{X})$. A subscheme $X \subseteq P^{r}_{A}$ is \textbf{projectively normal} for the given embedding, if its homogenous coordinate ring is an integral closed domain. Now assume that $k$ is an algebraic field, and that $X$ is a connected normal closed subscheme of $P^{r}_{k}$. Show that for some $d>0$, the $d-$uple embedding of $X$ is projectively normal.

Proof:

Proof: \textbf{ Claim I}: $S$ is a domain.\\ It is easy!\\

Let $S_{x_{i}}, 0\leq i\leq r,$ be a localization of $S$. It is a graded ring. For any element $x\in S_{x_{i}}$ we define the degree of x, $deg(x)$, by the degree of the lowest homogeneous part of $x.$ Construct a ring $\Gamma=\{x|x\in \cap_{0\leq i\leq r}S_{x_{i}}, deg(x)\geq 0\}$,

$\Gamma\subseteq Q(S) $ a subring of the quotient filed of $S$. Obviously, we have $S\subseteq \Gamma$. \\

\textbf{ Claim II}: $\Gamma$ is integral over $S$, and $\Gamma_{\geq n}=S_{\geq n}$ for sufficiently large enough $n$.\\

Let $y\in \Gamma.$ Then by the definition of $\Gamma,$ we have for any $x_{i}, 0\leq i\leq r,$ there exists an integer $n_{i}$, such that $x_{i}^{n_{i}}y\in S.$ Thus there is an integer $N,$ such that $ yS_{\geq N}\subseteq S.$ Especially, we have $x_{0}^{N}y\in S.$ On the other hand, since the degree of $y,$ $deg(y)\geq 0,$ we have for any integer $r$, $x^{N}_{0}y^{r}\in S.$ Thus we have $S[y]\subseteq S\frac{1}{x_{0}^{N}},$ where $S\frac{1}{x_{0}^{N}}$ is a finite generated $S$ module. So $y$, and furthermore the whole $\Gamma$ is integral over $S.$\\

Further, since $S$ is a finite generated domain over a filed $k$, we have that $\Gamma$ as a $S$ module, must be finite. And from the method we have used above, we could prove that for any $y\in \Gamma,$ there is an integer $N$, such that $yS_{\geq N}\subseteq S.$ Thus, we get for sufficiently large enough $n,$ $\Gamma_{\geq n}=S_{\geq n}.$ \\

Construct rings $^{i}\Gamma=\{x|x\in S_{x_{i}}, deg(x)\geq 0\}$, $0\leq i\leq r.$

Of course, we have $\Gamma=\cap_{0\leq i\leq r} {^{i}\Gamma}.$\\

On the other hand, it is easy to see that $^{i}\Gamma$ equals to the ring $S_{(x_{i})}[x_{i}]$, which are polynomial rings over $S_{(x_{i})}$, respectively. Since $S_{(x_{i})}$ are integral closed domains, so $^{i}\Gamma$, and further more $\Gamma$ are also integral closed domains. \\

Since $k$ is algebraic closed, we have $\Gamma^{(d)}=S^{(d)}$ for sufficiently large enough $d$. \\

Now let $y\in Q(\Gamma^{(d)}),$ the quotient filed of $\Gamma^{(d)}$, which is integral over $\Gamma^{(d)}.$ Then since $\Gamma$ is integral closed, we have $y\Gamma.$ On the other hand, since $y$ is almost integral over $\Gamma^{(d)},$ it is easy to see that $y\in \Gamma^{(d)}.$ So we proved that for sufficiently large $d$, the -uple embedding $S^{(d)}$ is integral closed.--