If $S=A[T_0,\cdots,T_d]$, $X=\operatorname{Proj}S$, then $O_X(n)(X)=S_n$ if $n\geq0$, $0$ otherwise.

Question

Here, $A$ is just some ring, $d>0$

In the following, $O_X(n)=\widetilde{S(n)}$, and $S(n)$ is $S$ with the graduation shifted (i.e. $S(n)_d=S_{n+d})$

and $S_{(f)}$ are elements of degree $0$ of $S_f$, i.e. elements of the form $\frac{P}{f^k}$ with $\operatorname{deg}P=k\operatorname{deg}f$

I have a proof I don't understand of this fact.

First I don't get why I can't just say that $O_X(n)(X)=\widetilde{S(n)}(D_+(1))=S(n)_{(1)}=S_n$ for all $n$ (and in particular, $S_n=0$ for negative degree since I assume the convention is that the only polynomial of negative degree is $0$, and we would be done. The proof I read is more involved and I don't get it.

Secondly, the proof is the following:

Let $B=A[T_0,\cdots,T_d,T_0^{-1},\cdots,T_d^{-1}]$, then a global section of $O_X(n)(X)$ is an element $f$ of $T_1^nO_X(D_+(T_1))$, so $f$ is of the form $P/T_1^m$

Why?

Then it concludes using the same argument with $T_0$ to deduce that $f$ has to be some homogeneous polynomial of degree $n$.

I understand that a section of $O_X(n)(X)$ restricted to $D_+(T_1)$ yields something of the form $\frac{P}{T_1^k}\in S(n)_{(T_1)}$ with $\operatorname{deg'}P=k\operatorname{deg'}T_1$, but here $P$ has degree $\operatorname{deg'}P=\operatorname{deg}P-n$ because of the shift, I mean, $P$ lies in $S(n)_{\operatorname{deg}P-n}=S_{\operatorname{deg}P}$.

If for some reason, $T_1$ still has degree $1$ then the argument in the proof makes sense in that we must have $k=\operatorname{deg}P-n$ to compensate $\operatorname{deg'}P$, so $\frac{P}{T_1^k}=T_1^n\frac{P}{T_1^{\operatorname{deg}P}}\in T_1^nO_X(D_+(T_1))$ but if $P$ has its degree shifted in $S(n)$, why wouldn't it be the same for $T_1$? Why isn't the degree of $T_1$ $1-n$? If it is the case then the argument does not make sense, we would need $\operatorname{deg}P-n=k(1-n)$ and I don't even know if such a $k$ exists now. Could somebody please clear that up?

Answer 1

First I don't get why I can't just say that $O_X(n)(X)=\widetilde{S(n)}(D_+(1))=S(n)_{(1)}=S_n$ for all $n$ (and in particular, $S_n=0$ for negative degree since I assume the convention is that the only polynomial of negative degree is $0$, and we would be done. The proof I read is more involved and I don't get it.

The statement that for $M$ a graded module over a graded ring $S$, the sections of $\widetilde{M}$ on $\operatorname{Proj} S$ over the open set $D(f)$ are equal to $M_{(f)}$ is only guaranteed by the construction of $\operatorname{Proj}$ when $f$ is of homogeneous of positive degree. $1$ is not of positive degree.

As for your confusions on the provided proof, let's recall a few definitions first so that we can be sure we're all on the same page.

Definition: Let $S$ be a graded ring and $M$ a graded $S$-module. Then for $f\in S$ homogeneous of positive degree, the sections of $\widetilde{M}$ on the open set $D(f)$ are exactly $M_{(f)}$, the degree-zero elements of the localization of $M$ at the multiplicative set $\{f,f^2,f^3,\cdots\}$.

In particular, this means that if $m\in M_d$ and $f\in S_1$, then $\frac{m}{f^d}$ is an element of $M_{(f)}$.

Now we get to your trouble. You write:

I understand that a section of $O_X(n)(X)$ restricted to $D_+(T_1)$ yields something of the form $\frac{P}{T_1^k}\in S(n)_{(T_1)}$ with $\operatorname{deg'}P=k\operatorname{deg'}T_1$, but here $P$ has degree $\operatorname{deg'}P=\operatorname{deg}P-n$ because of the shift, I mean, $P$ lies in $S(n)_{\operatorname{deg}P-n}=S_{\operatorname{deg}P}$.

If for some reason, $T_1$ still has degree $1$ then the argument in the proof makes sense in that we must have $k=\operatorname{deg}P-n$ to compensate $\operatorname{deg'}P$, so $\frac{P}{T_1^k}=T_1^n\frac{P}{T_1^{\operatorname{deg}P}}\in T_1^nO_X(D_+(T_1))$ but if $P$ has its degree shifted in $S(n)$, why wouldn't it be the same for $T_1$? Why isn't the degree of $T_1$ $1-n$? If it is the case then the argument does not make sense, we would need $\operatorname{deg}P-n=k(1-n)$ and I don't even know if such a $k$ exists now. Could somebody please clear that up?

Here it is important to remember that $T_1$ has in general different degrees considered as an element of $S(n)$ versus $S$. We know $T_1\in S_1$, which is the degree-$1$ piece of $S$, but the degree $1-n$ part of $S(n)$. This confusion between the degree of $T_1$ in the ring and the module appears to be a major source of your issues. You also appear to have a strange idea about what $\deg P$ ought to be.

Let's refer back to the comment after the definition and get things correct. The statement $\frac{P}{T_1^k}\in S(n)_{(T_1)}$ means that we should have $P\in S(n)_k$, which is to say that $P\in S_{n+k}$ by the definition of $S(n)$. So $P$, viewed as an element of $S$, is a homogeneous polynomial of degree $n+k$.