# Coordinate rings in projective spaces. What are they?

When $$X$$ is an algebraic variety of affine $$n$$-space, then the coordinate ring of $$X$$ are polynomials restricted to $$X$$.

But when $$X$$ is a variety in the projective $$n$$ space, what are the elements of its coordinate ring?

I know that the coordinate ring for the projective space is given by $$K[X_0,X_1,\dots,X_n]/I(X)$$, where $$I(X)$$ is a homogeneous ideal. But what does elements in $$K[X_0,X_1,\dots,X_n]/I(X)$$ look like? Are they homogemeous polynomials in $$K[X_0,X_1,\dots,X_n]$$ restricted to $$X$$ or are they all polynomials $$F$$ in $$K[X_0,X_1,\dots,X_n]$$ restricted to $$X$$, with $$F(la_0,la_1,\dots,la_n)=F(a_0,a_1,\dots,a_n)$$ for all $$l$$ in algebrically closed field $$K$$ and $$(a_0,a_1,\dots,a_n)$$ belongs to $$X$$?

• I don't think that you can interpret the elements of $K[X_0,X_1,\ldots,X_n]/I(X)$ as functions on $X$. You can interpret the homogeneous parts of this graded ring as global sections of a certain sheaf, though Commented May 10, 2013 at 5:55
• What do you mean by "look like"? Commented May 10, 2013 at 6:02
• By 'look like' i mean some simpler explanations of the elements. Commented May 10, 2013 at 6:07
• I want to get rid of the 'mod' thing. Commented May 10, 2013 at 6:43

The elements of $S=K[X_0,X_1,\dots,X_n]/I(X)$ are perfectly good functions...but not on $X$ !
They are functions on the cone $c(X)\subset \mathbb A^{n+1}$ associated to $X$, which has as equations the very elements of $I(X)$.
For example if $X$ is the circle $X_0^2+X_1^2-X_2^2=0$ in $\mathbb P^2$, then $c(X)$ is the cone $X_0^2+X_1^2-X_2^2=0$ in $\mathbb A^3$.
Unfortunately the elements of $S$ are not constant on the generatrices of the cone, so that the functions in $S$ do not descend to $X$, just as Jyrki and Matt told you.
You can experiment with $\overline X_0 \in S$ in the example above: $\overline X_0$ takes all values in $K$ on the generatrix $K(1,0,1)\subset c(X)$, so that it does not make sense to attach a value to $\overline X_o$ at the point $[1:0:1]\in X\subset \mathbb P^2$.

So the graded ring $S$ is useless, right? Wrong!
It is a ring with which one can build other interesting rings .
For example the field of rational functions $Rat(X)$ is the subfield of the fraction field $Frac(S)$ consisting of quotients $\overline P/\overline Q$ with $P,Q$ polynomials of the same degree and $Q\notin I(X)$.
Similarly one can define the local ring $\mathcal O_{X,x}$ for $x\in X$ by adding the condition $Q(x)\neq 0$.
[You might look at this answer for the case $X=\mathbb P^n$ ]
By the way, note that even if you cannot attach a value to $Q(x)$, it makes perfectly good sense to say that $Q(x)=0$ or $Q(x)\neq 0$.
And this is the source of more usefulness of the ring $S$: it permits you to define all subvarieties of $X$ by setting a bunch of homogeneous elements of $S$ equal to zero.

To sum up, consideration of the graded ring $S$ allows you to cut loose from the ambient $\mathbb P^n$ and be more intrinsic.
And this is the beginning of the Proj construction, which is the generalization to scheme theory of the classical notion of projective variety.

Edit:caveat
After all this praise of $S$, I must draw the attention to one drawback : it does not depend only on $X$ but also on its embedding into $\mathbb P^n$:
If for example we embed $\mathbb P^1$ into $\mathbb P^2$ successively as the line $X_0=0$ and then as the circle $X_0^2+X_1^2-X_2^2=0$ already mentioned , the corresponding graded rings are $S_1=K[X_1,X_2]$ and $S_2=K[X_0,X_1,X_2]/(X_0^2+X_1^2-X_2^2)$ and they are not isomorphic.

The elements of the coordinate ring are sections of certain line bundles on $\mathbb P^n$ that are restricted to $X$. Unfortunately, there is not really a simpler answer than that, because they are not functions on $X$, just as homo. polys. are not functions on $\mathbb P^n$.

• Ok. Its enough for me. Commented May 10, 2013 at 7:08
• Could the person who downvoted this fine answer by a remarkable mathematician please explain that strange decision? (I have upvoted it, of course, as usual when I read Matt) Commented May 10, 2013 at 10:23
• @GeorgesElencwajg: Dear Georges, Thanks for the compliments, but I understand the down-vote; it was a fairly abreviated answer. (These days I've noticed that questions can scroll down into obvlivion pretty quickly, so I thought I'd try the strategy of leaving a short answer when I had the chance, and come back later if the OP asked for more details.) Best wishes, Commented May 10, 2013 at 12:16