6
$\begingroup$

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?

Thanks.

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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? $\endgroup$
    – M Turgeon
    Commented Oct 5, 2012 at 14:29
  • $\begingroup$ What does 'Proj' stand for? What is Ex.2.14d? $\endgroup$
    – Berci
    Commented Oct 5, 2012 at 14:32
  • $\begingroup$ @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. $\endgroup$
    – Manos
    Commented Oct 5, 2012 at 14:36
  • 1
    $\begingroup$ @Berci: The Proj operator is described e.g. here : en.wikipedia.org/wiki/Proj_construction. I mention the reference to Exercise 2.14(d) for completeness, reproducing it here i think it would be confusing. $\endgroup$
    – Manos
    Commented Oct 5, 2012 at 14:42
  • $\begingroup$ @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. $\endgroup$
    – M Turgeon
    Commented Oct 5, 2012 at 15:57

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
17
  • 1
    $\begingroup$ No worries. Well what is the homogeneous coordinate ring of the variety $\mathbb{P}^{n}$. $\endgroup$
    – acyrl
    Commented Oct 5, 2012 at 15:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – acyrl
    Commented Oct 5, 2012 at 15:58
  • 1
    $\begingroup$ $\Bbb P^n$ as a scheme has many non-closed points. The notation $\{P\}^{-}$ usually means the Zariski closure of the point $P.$ $\endgroup$
    – Andrew
    Commented Oct 5, 2012 at 16:42
  • 1
    $\begingroup$ @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). $\endgroup$ Commented Oct 5, 2012 at 16:46
  • 1
    $\begingroup$ @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,... $\endgroup$ Commented Oct 5, 2012 at 18:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .