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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. –  user27126 Nov 22 '12 at 10:41
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? –  Tom Nov 22 '12 at 11:03
Oh, instead of $W$, should we take $\bar{W}$? –  Tom Nov 22 '12 at 11:40
Why is the map $p\mapsto Z$(points to irreducible closed sets) injective? –  Tom Nov 22 '12 at 11:57

2 Answers 2

up vote 4 down vote accepted

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. –  Tom Nov 23 '12 at 11:39

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? –  Tom Nov 22 '12 at 12:32
Yes, the correspondence I answered above is the natural generalization of the Nullstellensatz, because sheaves are the "global" version of rings. –  Fredrik Meyer Nov 22 '12 at 12:41
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. –  Tom Nov 22 '12 at 23:24
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 en.wikipedia.org/wiki/Sober_space –  user27126 Nov 23 '12 at 7:27
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 ) –  Georges Elencwajg Nov 23 '12 at 8:33

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