Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Hartshorne's "Algebraic Geometry" p. 77, Example 2.5.1, it is mentioned that if "$k$ is an algebraically closed field, then the subspace of closed points of $\operatorname{Proj} \, k[x_0,\cdots,x_n]$ is naturally homeomorphic to the projective $n$-space $\mathbb{P}^n$. He refers to Ex. 2.14d, however I don't see the connection. Any insights?


P.S. Ex. 2.14(d) seems to me a little bit obscure at this point, this is why i am not reproducing it. Any argument relating $\operatorname{Proj} \, k[x_0,\cdots,x_n]$ and $\mathbb{P}^n$ is very welcome.

share|cite|improve this question
Do you understand the connection between $Spec(k[x_1,\ldots,x_n])$ and $\mathbb A_k^n$, when $k$ is an algebraically closed field? – M Turgeon Oct 5 '12 at 14:29
What does 'Proj' stand for? What is Ex.2.14d? – Berci Oct 5 '12 at 14:32
@MTurgeon: Very good question. No, i don't. I understand the connection between $\operatorname{Specm}(k[x_1,\cdots,x_n])$ and $\mathbb{A}^n_k$. Its a $1-1$ correspondence. If you can point to me the relevant theorems in Hartshorne i would appreciate it. – Manos Oct 5 '12 at 14:36
@Berci: The Proj operator is described e.g. here : I mention the reference to Exercise 2.14(d) for completeness, reproducing it here i think it would be confusing. – Manos Oct 5 '12 at 14:42
@Manos As noted below by acyrl, you could have a look at The Geometry of Schemes by Eisenbud-Harris; the relevant section is II.1.1. Also, this is discussed in Hartshorne. Have a look at Proposition 2.6 in Chapter 2. – M Turgeon Oct 5 '12 at 15:57
up vote 5 down vote accepted

Exercise $2.14~ d)$ states that for any projective variety with homogeneous coordinate ring $S$, $$t(V) \simeq \operatorname{Proj}S$$ Which include $\mathbb{P}^{n}$, meaning $V$ could be $\mathbb{P}^{n}$.

Now by proposition 2.6, $V$ and $t(V)$ have homeomorphic closed points.

share|cite|improve this answer
No worries. Well what is the homogeneous coordinate ring of the variety $\mathbb{P}^{n}$. – acyrl Oct 5 '12 at 15:23
Do you mean $\mathbb{P}^{n}$ as in the variety sense or the scheme. In the variety sense they are all closed, but not in the scheme $\operatorname{Proj} S$ sense. – acyrl Oct 5 '12 at 15:58
$\Bbb P^n$ as a scheme has many non-closed points. The notation $\{P\}^{-}$ usually means the Zariski closure of the point $P.$ – Andrew Oct 5 '12 at 16:42
@Manos: you have to be careful by what you mean by "variety". $\operatorname{Proj}S$ is a scheme and has not the slightest chance to be a variety in the sense Hartshorne defines varieties in Chapter 1. However Proposition 2.6 shows that every such "old-fashioned" variety gives rise to a scheme in a canonical way. So the new definition of (abstract) variety should be that it is a scheme, which is isomorphic to a scheme, which is in the image of the functor of 2.6 (but see 4.10 for a more concrete characterization). – Nils Matthes Oct 5 '12 at 16:46
@Manos: 1) Yes, $\mathbb{P}^{n}$ is irreducible (as its coordinate ring is an integral domain). 2) I think I didn't make my point clear enough, thanks for being persistent. A variety in the sense of chapter one does always live in an ambient space, which is $\mathbb{P}^{n}$ for some $n$ depending on the variety. So there really is no relation between $\operatorname{Proj}k[x_0,...,x_n]$ (which is an "abstract" topological space) and $\mathbb{P}^{n}$ as a variety in the sense of Ch.1 a priori. Also $\operatorname{Proj}k[x_0,...,x_n]$ will always contain a non-closed point,... – Nils Matthes Oct 5 '12 at 18:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.