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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.

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    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?2012-10-05
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    What does 'Proj' stand for? What is Ex.2.14d?2012-10-05
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    @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.2012-10-05
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    @Berci: The Proj operator is described e.g. here : http://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.2012-10-05
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    @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 2.2012-10-05
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    @MTurgeon: Thanks. I've been studying proposition 2.6 since the day before yesterday, it takes some time to digest :)2012-10-05

1 Answers 1

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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.

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    Thanks for the answer. How does $k[x_0,\cdots,x_n]$ come into play?2012-10-05
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    No worries. Well what is the homogeneous coordinate ring of the variety $\mathbb{P}^{n}$.2012-10-05
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    It is $k[x_0,\cdots,x_n]$. So in general, $\mathbb{P}^n$ is different from $\operatorname{Proj} k[x_0,\cdots,x_n]$. Is there a sense (or conditions) in which we have equality?2012-10-05
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    In the sense above. That they have the "same" closed points, but $\operatorname{Proj} S$ has points which are not close. As mentioned above, you can gain some intuition by thinking about the affine case. Geometry of Schemes by Eisenbud has quite a few examples with pictures that can help you out. Its worth noting that sometimes the notation $\mathbb{P}^{n}$ really means $\operatorname{Proj} S$, when you are exclusively talking of schemes.2012-10-05
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    Thanks. All points of $\mathbb{P}^n$ are closed?2012-10-05
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    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.2012-10-05
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    Yes, i meant in the variety sense. So is it accurate to say that $\operatorname{Proj}S$ is a projective variety if and only if all of its points are closed? And one more short question regarding Proposition 2.6: in the proof of 2.6 what is the notation $\left\{P\right\}^-$ in defining the map $\alpha$ in the end of the first paragraph?2012-10-05
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    $\Bbb P^n$ as a scheme has many non-closed points. The notation $\{P\}^{-}$ usually means the Zariski closure of the point $P.$2012-10-05
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    @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).2012-10-05
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    @NilsMatthes: Thanks for your comment. Hartshorne requires any variety to be irreducible. He also refers to $\mathbb{P}^n$ as projective variety. 1) Is $\mathbb{P}^n$ irreducible? 2) Since every point of $\mathbb{P}^n$ is closed in the variety sense, if all points of $\operatorname{k[x_0,\cdots,x_n]}$ are closed, then the latter is topologically isomorphic to $\mathbb{P}^n$. Correct?2012-10-05
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    @NilsMatthes: I also encountered in some notes the statement "$\operatorname{Proj} k[x_0,\cdots,x_n]$ is isomorphic to the algebraic $\mathbb{P}^n$." Any idea what that means?2012-10-05
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    @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,...2012-10-05
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    ...corresponding to the zero ideal of $k[x_0,...,x_n]$, so your statement that $\operatorname{Proj}k[x_0,...,x_n]$ consists of closed points only is not true (except for $n=0$). As for your second comment, in a scheme-theoretic setting, ignoring "old-fashioned" varieties one usually defines $\mathbb{P}_{k}^{n}$ as $\operatorname{Proj}k[x_0,...,x_n]$. You may even replace $k$ by an arbitrary commutative ring, so with schemes at hand, it makes even sense to talk about, e.g. "projective $n$-space over $\mathbb{Z}$". But it's hard judging the content of your 2nd comment w/o knowing its source2012-10-05
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    @NilsMatthes: Thank you for your very enlightening comments. So, since $\operatorname{Proj} k[x_0,\cdots,x_n]$ always contains a non-closed point (for n>0), and any point of a projective variety is closed, we can say that $\operatorname{Proj} k[x_0,\cdots,x_n]$ is a projective variety if and only if $n=0$.2012-10-05
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    @Manos: You're right, $\operatorname{Proj}k[x_0]$ has only one point, and hence is isomorphic to a projective variety, namely to any one-point variety in $\mathbb{P}^{n}$. We should better stop discussing here, though, as discussions in the comments are commonly discouraged on this site. Maybe you want to digest the information you gained and ask some more questions later?2012-10-05
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    @NilsMatthes: Definitely, i was planning to do that :) Thanks again for your help.2012-10-05
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    Sorry, I had gone away from the computer and missed the follow up comments. Glad someone else helped. =)2012-10-07