# Notation in Hartshorne Exercise 1.2.6

I am now doing Hartshorne Problem 1.2.6.

Hartshorne 1.2.6: Let $Y$ be a projective variety with homogeneous coordinate ring $S(Y)$, show that $\dim S(Y) = \dim Y + 1$. [Hint: Let $\varphi_i : U_i \to \Bbb{A}^n$ be the homeomorphism of (2.2), let $Y_i$ be the affine variety $\varphi(Y \cap U_i)$ be its affine coordinate ring. Show that $A(Y_i)$ can be identified with the subring of elements of degree $0$ of the localized ring $S(Y)_{x_i}$. Then show that $S(Y)_{x_i} \cong A(Y_i)[x_i,x_i^{-1}]$.

Now I have done most of the exercise but have several questions related to it:

1. The notation $S(Y)_{x_i}$ means localization at the multiplicative set $\{1,x_i,x_i^2,\ldots\}$ yes? It does not mean localization at the prime ideal $(x_i)$ right? I don't think the two coincide.

2. Assuming I understand the notation right, I am confused how we can just do $S(Y)_{x_i}$ without some extra assumptions. In particular it could be that $x_i$ is zero in $S(Y)$, i.e. $x_i \in I(Y)$ then we are in trouble because the localization is the zero ring. Should there be an extra assumption here?

3. Also how can I say that $S(Y)_{x_i} \cong A(Y_i)[x_i,x_i^{-1}]$? I know how to prove this if $S(Y)$ is a $\Bbb{Z}$ - graded ring but now it is just a $\Bbb{N}$ - graded ring.

1. Yes, $$S(Y)_{x_i}$$ means localization at the multiplicative set $$\{1,x_i,x_i^2,\ldots\}$$ .
Not only does it not coincide with $$S(Y)_{(x_i) }$$ but in a sense it is complementary, as seen in the following affine example:
Let $$\mathbb A^1_k=\text{Spec}(k[T])$$. Then $$k[T]_T$$ is the ring of regular functions on $$\mathbb A^1_k\setminus \{0\}$$,i.e. defined outside $$0\in \mathbb A^1_k$$, whereas $$k[T]_{(T)}$$ is the ring of germs of functions defined only near $$0\in \mathbb A^1_k$$ .

2. You are very attentive: congratulations!
But Hartshorne's isomorphism is correct: if $$x_i\in I(Y)$$, then $$Y$$ is contained in the hyperplane $$H_i\subset \mathbb P^n_k$$ given by $$x_i=0$$.
Since $$Y\subset H_i=\mathbb P^n_k\setminus U_i$$ , its intersection with $$U_i$$ is empty: $$Y_i=Y\cap U_i=\emptyset$$.
The ring of functions of $$Y_i$$ is thus zero: $$A(Y_i)=0$$, and since $$S(Y)_{x_i}=0$$ too because you are inverting $$x_i=0\in S(Y)$$ Hartshorne's isomorphisms reduces to $$0\cong 0 :$$ not exciting but true!

3. As you correctly write in your comment below, you may use that for a $$\mathbb Z$$-graded ring $$R$$ and a homogeneous element $$f\in R_1$$ of degree $$1$$ you have the isomorphism of $$\mathbb Z$$-graded rings $$(R[f^{-1}]_0)[x,x^{-1}]\stackrel {\cong}{\to} R[f^{-1}]:\sum_{j\in \mathbb Z} q_jx^j\mapsto \sum_{j\in \mathbb Z} q_j f^j \quad (q_j\in R[f^{-1}]_0)$$
You would like to apply this to $$R=S(Y)$$ and $$f=x_i$$, but there is the apparent difficulty that $$S(Y)$$ is $$\mathbb N$$-graded instead of being $$\mathbb Z$$-graded.
The solution is purely formal: just decree that $$S(Y)_n=0$$ for $$n\lt 0$$ and $$S(Y)$$ becomes a $$\mathbb Z$$- graded ring.
This is a general formal procedure allowing to consider any $$\mathbb N$$- graded ring as a $$\mathbb Z$$- graded ring.

• Dear Georges, I was waiting so badly for you to answer, thanks so much! For 3, it was a typo above that I forgot to say $S(Y)_{x_i}$. Now the thing with this exercise for me is, if a priori $S(Y)$ is a $\Bbb{Z}$ - graded ring then I conclude the isomorphism from this general fact from commutative algebra: Let $R$ be a $\Bbb{Z}$ - graded ring, $f \in R$ not zero and of degree $1$. Then $R_f \cong (R_f)_0[x,x^{-1}]$ (Exercise 2.17, Eisenbud). But my problem now is like I said I can't apply this exercise because $S(Y)$ is only a priori $\Bbb{N}$ - graded.
– user38268
Commented Apr 21, 2013 at 7:58
• Dear Benja, I have modified part 3. of my answer, deleting the allusion to your now corrected typo and addressing the problem of $\mathbb N$-grading. Commented Apr 21, 2013 at 8:45
• Dear Georges, I can't believe the solution to the problem I outlined in (3) was so simple! Sometimes I think I'm blind. I think I can complete this exercise except I still need to think out how the isomorphism above helps in computing the dimension of $S(Y)$. I will speak to my advisor tomorrow and post a question if I still have one.
– user38268
Commented Apr 21, 2013 at 9:06
• Dear Benja, you do well to dissect that exercise in every detail. Ideally, the exercise should be amplified to a whole section serving as an introduction to the Proj construction , which is fundamental in scheme theory. Unfortunately Hartshorne does not do it for lack of space in his already thick book , nor do most (all?) other authors. The unfortunate consequence is that this Proj construction is hopelessly abstract and quite difficult to make sense of the first time one sees it. Fortunately, for you Hartshorne's presentation on page 76 of Proj will now be far more accessible . Commented Apr 21, 2013 at 10:13