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Can anybody help me in understanding the hint given in the problem $2.6$ in Chapter $1$ of Hartshorne's Algebraic Geometry book ?

I cannot see why $A(Y_i) $ can be identified with degree zero elements of $S(Y)_{x_i}$. Actually I cannot find it out how does the degree zero elements in $S(Y)_{x_i}$ look like.

Please Help me. Thanks.

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    Hint: If $x_0,\dots x_n$ are coordinates on $\mathbb{P}^n$, then the coordinates on $U_i\cong \mathbb{A}^n$ are $\frac{x_0}{x_i},\dots, \frac{x_n}{x_i}$. This means $A(Y_i)$ are regular functions on $Y_i$ in the $x_j/x_i$, $j=1,\dots, n$. But these are precisely the elements of degree $0$ in $S(Y)_{x_i}$. – Andrew May 05 '16 at 14:37
  • I cannot really understand arguments by regular functions as it is defined in the next section and also how the elements in $S(Y)_{x_i}$ looks like. I know that $S(Y) $ is the homogeneous coordinate ring of the projective variety Y, i.e., quotient of a polynomial ring in (n+1) variables over K by a prime ideal. All I want to clarify my ambugities. –  May 05 '16 at 15:41

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We know that $Y_{i}=Y\cap A_{i}$ cover $Y$. By definition of the dimension we have $\dim(Y)=\sup \dim (Y_{i})$. Now in local coordinates we have $$ A[Y_{i}]=S[Y]_{(x_{i})}|^{0} $$ as Andrew pointed out. To see this let us write $S(Y)=\oplus_{d\ge 0} S_{d}/I(Y)$. For any function $f \in S(Y)$, its affine piece on $A_{i}$ is given by $$ f(\frac{x_{1}}{x_{i}},\cdots, \frac{x_{n}}{x_{i}}) $$ and since each term $\frac{x_{1}}{x_{i}}$ has degree $0$, $f$ itself must be of degree $0$ as well. Thus every element in $A[Y_{i}]\subset S[Y]_{(x_{i})}|^{0}$, and the reverse inclusion is not difficult. Now we have $$ S[Y]_{(x_{i})}=A[Y_{i}][x_{i},x_{i}^{-1}]\rightarrow \dim(S[Y]_{(x_{i})})=\dim(A[Y_{i}])+1 $$ And since localization preserves chains of prime ideals, we have $$\dim(S[Y])\ge \dim(S[Y]_{(x_{i})})=\dim(A[Y_{i}])+1$$ Thus $$\dim(S[Y])\ge \sup \dim(Y_{i})+1=\dim(Y)+1$$ And we know $\dim(S[Y])=\dim(S[Y]_{(x_{i})})$(Why?). So we conclude $\dim(S[Y])=\dim(Y)+1$.Mind that the above argument trivially collapses when $Y\cap A_{i}=\emptyset$ or $x_{i}\in I(Y)$, etc. So some extra care is needed.

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Bombyx mori
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  • Few doubts as i start reading it: what is S[Y]_(xi) ? –  May 08 '16 at 10:18
  • Maybe better written as $S[Y]{x{i}}$? But the later notion did not really show $(x_{i})$ as a multiplicative set. – Bombyx mori May 08 '16 at 10:22
  • Can you show me a typical element in the degree 0 element in $S[Y]{x_i}$ where S[Y]= S/I(Y) as in Hartshornes notation. I understand that $S[Y]{x_i}$ means localization of the ring $S/I(Y)$ with respect to the image of the multiplicative set {1,$x{_i}$,$x{_i}^2$,....} –  May 08 '16 at 10:34
  • Sure. I think $\frac{x_{j}}{x_{i}}$ would be such an element if $x_{j}\not\in I(Y)$. – Bombyx mori May 08 '16 at 10:35
  • I cannot see it. What I know that If $R$ is a $N$-graded ring and $I$ is a homogeneous ideal of it and if $S$ multiplicative closed set of $R$ consists of only homogeneous elements that does not intersect $I$ then $(R/I)_{S}$ becomes $Z$-graded. This set up also match with this problem. Can you elaborate above in this set up? –  May 08 '16 at 10:45
  • I already tried my best to elaborate and gave you a simple example above. I do not see your confusion - you cited a lot of definition but could not pinpoint anything specific. – Bombyx mori May 08 '16 at 10:47
  • Just explain me why $x_{j}/x_{i}$ is an element in $S[Y]{x{i}}$ as you gave me an example . –  May 08 '16 at 10:52
  • This is already the simplest example possible. If you do not get it, I think you need to review $\text{Proj}$ construction earlier in the book. Maybe you can try some other easier texts too (like Harris' book). – Bombyx mori May 08 '16 at 10:54
  • One more thing in the 4th line of your answer will it not be $I_{d}(Y)$ in the denominator inside the direct sum ? –  May 08 '16 at 11:00
  • You do not have to accept my answer if you do not get it. Let me see if I argued in a confusing way. Let $Y=(t,t^2,t^3)$ be the twisted cubic in the book. Then it can be written as $k[x,y,z,w]/(xz-y^2, zw-xy, x^2-wy)$ in the projective plane $\mathbb{P}^{3}$. Now given $Y'\in \mathbb{P}^{n}$ cut out by these pieces, you just need to let $w=1$ to recover the original affine variety. – Bombyx mori May 08 '16 at 11:01
  • You can try to view $S=\oplus S_{d}$ as a ring (of homogeneous degree elements) itself, and $I(Y)=\oplus I_{d}$ as well. So write $S[Y]=\oplus S_{d}/I(Y)$ is really just a short hand way of writing out the quotient. – Bombyx mori May 08 '16 at 11:03
  • I am confused with only one thing : what will be the degree 0 component of localization of a quotient ring ? –  May 08 '16 at 11:05
  • Let us consider $S=\oplus S_{d}$, where $S_{d}$ are homogeneous polynomials in degree $d$ in $x,y$. Now if we localize at ${1,y,y^2,\cdots }$, then the degree zero piece is the $\textbf{total degree}$ zero piece in the localized ring viewed as a subring of $K(S)=k(x,y)$. So for example $\frac{x^4}{y^2}$ will have degree 2, and hence not in the degree $0$ piece. – Bombyx mori May 08 '16 at 11:10
  • Thanks @Bombyx morifor for giving me enough material to convince myself. –  May 08 '16 at 11:11
  • I think Ravi Vakil's notes did this much more carefully, if you are interested to read it. The "core observation" you need to focus is that to projectify an affine variety, you substitute $f(x_1,\cdots x_{n})$ by $f(x_{1}/x_{i},\cdots x_{n}/x_{i})$, and then expand out the equation into polynomial form. To "get the affine piece", you did the reverse by simply letting $x_{i}=1$. Since we are working with projective coordinates this is okay. – Bombyx mori May 08 '16 at 11:15
  • So hopefully this clarifies the issue. If you are still confused, try to work out some examples (the quadratic surface, the twisted cubic, the Segre embedding, the $d$-tuple embedding, etc) and see how it corresponds in affine pieces. Maybe I am not great at elaborating mathematics. But there are masters you can learn from. For reference, Joe Harris's book and Ravi Vakil's notes are very good (just Google it). – Bombyx mori May 08 '16 at 11:19