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I know that there is a bijection between the irreducible closed sets and the prime ideals of a ring (via functions $V()$ and $I()$).

Is this true not only for affine schemes but also for general schemes? And Why?

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    There is a bijection between irreducible closed sets and points of scheme. This can be proved by checking on affine open sets.2012-11-22
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    Thank you. I understand that it holds in each affine open set. For any point $p$ of a scheme we can find an irreducible closed set $W$ in an affine open set containing $p$. But $W$ might not be closed in the global scheme?2012-11-22
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    Oh, instead of $W$, should we take $\bar{W}$?2012-11-22
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    Why is the map $p\mapsto Z$(points to irreducible closed sets) injective?2012-11-22

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Response to the comment:

The correspondence is a point $p$ to $\overline{\{p\}}$, the closure of $p$. It says two things: existence (that each irreducible closed set has a generic point), and uniqueness (that each irreducible closed set has a unique generic point)

Let's say our scheme is $X$, an irreducible closed set being $C$.

Existence: We want to find a generic point of $C$.

Intersect $C$ with an affine open subset $U$. This intersection is dense in $C$, while by the affine case $C \cap U$ has a generic point $p$ (considered as a subset of $U$) in $U$. Try to show that the closure of $p$ in $X$ is $C$.

Uniqueness: Suppose $p$ and $q$ has the same closure $C$. We want to show that $p = q$.

Take an affine open subset $U$ that contains $p$, and try to show that $p$ is the same as the generic point of $C \cap U$ in $U$.

Now take an affine open subset $V$ that contains $q$. Irreducibility of $C$ forces $U \cap V$ to be nonempty. The last paragraph then shows that both $p$ and $q$ are the generic point of $C \cap U \cap V$ in $U \cap V$, which means that they are the same.

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    I have just understood your answer! Thank you very much.2012-11-23
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    The last paragraph is not clear since $U\cap V$ might not be affine. Instead, one should argue that $p,q\in U\cap V\subseteq U$ and that their closures in $U$ are both $C\cap U$.2016-12-31
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This is the whole point of scheme theory. Hilbert's Nullstellensatz only gives a bijection between radical ideals in $k[x_1,\cdots,x_n]$ and closed subsets of $\mathbb{A}^n$, for $k$ algebraically closed.

Now for any scheme, there is 1-1 correspondence between quasi-coherent ideal sheaves (subsheaves of the structure sheaf) and closed subschemes of X. (this is the content of proposition 5.9 in Harthorne)

The correspondence is given as follows: If $Y$ is a closed subscheme of $X$, then the inclusion morphism $i:Y \to X$ gives a morphism of sheaves: $i^\#:\mathcal{O}_X \to i_*\mathcal{O}_Y$. The ideal sheaf is the kernel of this map.

Conversely, given a scheme $X$ and a quasi-coherent shraf of ideals $\mathcal{J}$, let $Y$ be the support of the quotient sheaf $\mathcal{O}_X/\mathcal{J}$. Then $Y$ is a closed subscheme with structure sheaf $\mathcal{O}_X/\mathcal{J}$.

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    Thanks. There are several versions of Nullstellensatz. One of which says that any (not-necessarily-alg.closed)ring $A$ has a bijection between the irreducible closed sets of Spec $A$ with Zariski topology and the prime ideals of $A$. Is this also true for schemes?2012-11-22
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    Yes, the correspondence I answered above is the natural generalization of the Nullstellensatz, because sheaves are the "global" version of rings.2012-11-22
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    Thank you. But I haven't yet learned the words coherent and closed-subscheme. Probably I will be able to read your answer two months later.2012-11-22
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    This answer actually does not capture the point. An irreducible closed SUBSET does not capture any information of functions (which means it has nothing to do with sheaf), and is something purely topological. To read more about these spaces, see http://en.wikipedia.org/wiki/Sober_space2012-11-23
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    Dear @Sanchez, your comment is unnecessarily harsh: Frederik's answer is an excellent answer to Tom's question and goes even further than just answering it, precisely because scheme theory takes not purely topological structures into account. If you (wrongly) insist on pure topology, just restrict Frederik's perfect 1-1 correspondence to *reduced* subschemes on one hand and to *radical* quasi-coherent ideals ($\mathcal I=\sqrt {\mathcal I}$) on the other (And, by the way, I know what a sober topological space is since I happen to be aware of the existence of [both versions of] EGA I )2012-11-23
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    @GeorgesElencwajg, I am sorry if I sounded rude in the previous comment - it was most definitely not my intention at all. I saw what I was proposing as a direct generalization of what OP was asking - so to me Fredrick's answer, while is perfectly written in its own right, seems to be slightly off track to me. (This is of course a highly subjective opinion.) Sorry again if what I wrote (without any harshness in mind to be honest) offended anyone.2012-11-23
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    Dear @Sanchez, that's quite all right: I essentially wanted to emphasize the quality of Frederik's answer. I should have added that your answer is interesting too: +1 for it!2012-11-23