I'm currently working on exercise 2.6 in chapter 1 of Algebraic Geometry by Hartshorne. I'm pretty confident with my answer apart from the first bit which I feel I could be "fudging". I'm looking for people to tell me where I have gone wrong... Here is the set up for those who don't have the book:

If $S=k[y_0, \dots ,y_n]$ is a graded ring, let $S_{(f)}$ denote the subring of $S_{f}$ consisting of elements of degree $0$. Let $Y$ be a projective variety embedded in $\mathbb{P}^{n}$ with homogeneous coordinate ring $S(Y)$, let $H$ be the hyperplane $y_0=0$, let $U=\mathbb{P}^{n}-H$, and let $X=Y\cap U$. The claim is that the affine coordinate ring $A(X)$ is isomorphic to $S(Y)_{(y_0)}$.

My answer with help from the book: define an isomorphism $\theta$ from $k[x_1, \dots ,x_n]$ to $S_{(y_0)}$ by mapping $f(x_1, \dots ,x_n)$ to $f(y_1/y_0, \dots ,y_n/y_0)$. The image of $I(X)$ under this map is equal to $I(Y)_{(y_0)}$. Indeed, if $f(x_1, \dots ,x_n)$ vanishes on $X$, then $y_0^{d}f(y_1/y_0, \dots ,y_n/y_0)$ vanishes on $Y$, and hence, $f(y_1/y_0, \dots ,y_n/y_0)=y_0^{d}f(y_1/y_0, \dots ,y_n/y_0)/y_0^{d}$ is an element of $I(Y)_{(y_0)}$. To show surjectivity, it is enough to there is an element of $I(X)$ which maps to each generator of $I(Y)_{(y_0)}$. If $F(y_0, \dots ,y_n)/y_0^{d}=y_0^{d}F(1,y_1/y_0 \dots ,y_n/y_0)/y_0^{d}$ is such a generator, then $F(x_1 \dots x_n)$ is an element of $I(X)$ which maps to it. Thus, $\theta$ induces an isomorphism from $A(X)$ to $S_{(y_0)}/I(Y)_{(y_0)}$, and the latter is clearly isomorphic to $S(Y)_{(y_0)}$.

Also, is there an easier way to see/think about this? For example, some books I have read have stated that $A(X)$ can be identified with $k[\xi_1/\xi_0 \dots \xi_n/\xi_0]$ (where the $\xi_i$ are the coordinate functions of $Y$) without further justification.

Thanks for any help

  • $\begingroup$ Dear M Davolo, If you're still interested in this: in your argument, what is the value of $d$? And why does $y_0^d f(y_1/y_0,\cdots,y_n/y_0)$ vanish on $Y$? Regards, $\endgroup$ – Matt E May 9 '13 at 4:18

I must admit i looked at your question only quickly, but i do believe there's an error: to show surjectivity, you must show that every element in $S_{(y_o)}$ has a preimage, not every element in $I(Y)_{(y_0)}$.

From your question i get that you're working your way through Hartshorne. In that case i think with your current knowledge your solution is good. However have a look at the following reasoning:

Oh by the way, on this page you can find solutions to many of the exercises in Hartshorne, http://www.math.northwestern.edu/~jcutrone/research.html, they helped me a lot.

The point you need to consider, is that $X$ is not immediately affine, at first it is just quasi projective. We need the well known map $$ \phi: U \longmapsto \mathbb{A}^n $$ which maps $[a_0:\ldots:a_n]$ to $(a_1/a_0,\ldots,a_n/a_o)$. This gives an isomorphism (check Hartshorne) of varieties, and gives $X$ the structure of an affine variety. It is only then that we can calculate its affine ideal and hence its coordinate ring $A(X)$.

Now here's comes a quick argument for your result that does require theory from some later chapters of Hartshorne..

By the map above, the quasi projective variety $Y\cap U$ is isomorphic to its image, lets say $Z$ (this is NOT the same thing as your $X$, although they are of course isomorphic). Then the rings of global regular functions are isomorphic $\mathcal{O}_Z(Z) \cong \mathcal{O}_X(X)$. Since $Z$ is affine, $\mathcal{O}_Z(Z) \cong A(Z)$, and by a theorem that Hartshorne unfortunately only states for schemes, (II 2.5), we have $\mathcal{O}_X(X) \cong S(Y)_{(y_0)}$, hence your result. The result is quite immediate, i believe that that is the reason some authors just state it as "known".

Let me know if you still have questions.

  • $\begingroup$ Thanks for your answer. Aside from when I first defined $X$, when I wrote $X$ I really meant to write $Z$ (using your notation), but was sloppy and forgot. With regards to the error, I wasn't trying to show that $\theta$ is surjective (this is obvious, right?!); I was trying to show that the image of $I(Z)$ under $\theta$ is $I(Y)_{(y_0)}$. In light of this, am I correct or is there still an error? $\endgroup$ – M Davolo Aug 14 '12 at 19:33
  • $\begingroup$ Thanks for your answer by the way - it seems a lot cleaner. Is there a way to prove that $\mathcal{O}_X(X)$ is isomorphic to $S(Y)_{(y_0)}$, outside of the theory of schemes or by the way I have attempted? $\endgroup$ – M Davolo Aug 14 '12 at 19:37
  • $\begingroup$ No prob. Ah i see, you want to show that the restricted map $I(X) \rightarrow I(Y)_{(y_0)}$ is surjective? This is not necessary. In fact there are isomorphic rings for which this does not hold. In general, if you want to show isomorphism of rings that might be quotients by some ideal, you'll need to show the following: (1) the morphism is well defined (here you need to show that the image of $I(X)$ is CONTAINED in $I(Y)_{(y_0)}, i.e. the map is independent of choice of representative (2) it is a ring morphism and (3) it is injective and surjective, or equivalently, construct an inverse. $\endgroup$ – Joachim Aug 14 '12 at 19:48
  • $\begingroup$ For the second question, i'm sure it will, in fact Hartshorne does it for affine varieties in 4.2. I can imagine this proof generalizing (although then i don't see why Hartshorne didn't). But also, using scheme theory for varieties is basically using another language for the same thing (although much more powerful), i'm sure anyone who knows schemes could translate the proof Hartshorne gives in II.2.5 back to the language of varieties. Trying to see if I.4.2 generalizes might be a good exercise for you later on. If i find time i will try it myself. $\endgroup$ – Joachim Aug 14 '12 at 19:57

Your proof looks good.

The core of the argument seems to me to be the relation between dehomogenization and homogenization of polynomials (or more precisely, of regular functions). $\theta$ "homogenizes" them (well not quite, it also divides by a suitable power of $y_{0}$ in order to get a regular function on $X$) and its inverse dehomogenizes a regular function on $X$ with respect to $y_{0}$ to obtain a polynomial in $n$ variables.

On a second note, I guess that the exercise was just meant to express the regular functions on your affine variety $X$ in terms of the homogenuous coordinate ring $S(Y)$ of the ambient variety $Y$ and in this light it is quite natural that you have to invert $y_{0}$ in $S(Y)$, because $y_{0}$ doesn't vanish on $X$, hence dividing by $y_{0}$ is okay.

With this, and some more work (as outlined in Hartshorne Ex. 2.6), you can relate the dimension of $S(Y)$ to the dimension of $Y$. To do so, you first reduce to the "local" (i.e. the affine) case, in which you already know the result (i.e. $\dim(X)=\dim(A(X))$), and then the "global" (projective) result follows more or less directly from your result, using Ex.1.10b). Arguments of that kind (reduction to the affine case and recovering the global result from it) appear frequently throughout algebraic geometry.

As for your second question, it would be helpful to know which books you mean.


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