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In the chapter I.3.3 of Eisenbud & Harris "The Geometry of Schemes" they give a definition of "prime ideal sheaf":

Let $\mathcal{O}_S$ be the structure sheaf of a scheme $S$, and let $\mathcal{F}$ be a quasicoherent sheaf of $\mathcal{O}_S$-algebras. Then a prime ideal sheaf $\mathcal{I} \subset \mathcal{F}$ is a quasicoherent sheaf of ideals of $\mathcal{F}$, such that on eash open affine $U \subset S$ the ideal $\mathcal{I}_U$ is prime (or unit).

After that they claim that prime ideal sheaves of $\mathcal{O}_S$ itself correspond to points of $S$. And here probably I miss some point, since I don't see why the following is not a counterexample.

Consider the following non-separated scheme, presented earlier in the same book. Let $S$ be a scheme that is obtained by gluing two copies of the affine line $S_1 \simeq S_2 \simeq \mathrm{Spec} \, k[X]$ along the complement of $0$ (I mean the ideal $(X)$, of course) by identity morphism, and let $0_1$ and $0_2$ be the "two zeroes" of $S$. Then I think that the sheaf of ideals corresponding to the reduced two-point subscheme supported on $\{0_1, 0_2\}$ satisfies the definition of prime ideal sheaf, since no affine open subset can contain both points simultaneously. However, I don't think this sheaf of ideals deserves to be called prime, and it clearly violates their claim.

So where am I wrong? Or if it is really a counterexample, what is the right definition then? Is it enough to add that $\mathcal{F}/\mathcal{I}$ should have irreducible support?

Edit: It turned out that in the (more recent) paper edition they only claim that for affine schemes.

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Where exactly do they claim that the prime ideal sheaves correspond to points of $S$? As far as I see, they only claim this for affine $S$. –  Vladimir Sotirov May 4 '13 at 21:46
    
@Vladimir: In my edition of the book, directly after the definition, it is remarked (in brackets!): "(Observe that for any scheme $X$, the points of $X$ are simply the prime ideal sheaves of $\mathcal{O}_X$.)" –  Martin Brandenburg May 4 '13 at 21:49
    
@VladimirSotirov: Moreover, the same is actually given as Exercise I-51. –  Alexander Shamov May 4 '13 at 21:56
    
In my edition it says "(Observe that for any affine scheme $X$, the points of $X$ are simply the prime ideal sheaves of $\mathcal O_X$.)" The exercise likewise asks to show that the points of an affine scheme correspond to prime ideal sheaves (restricted to every open affine it is a prime ideal or the unit ideal). –  Vladimir Sotirov May 4 '13 at 21:56
    
Mine is probably a draft compiled directly from latex, so I assume they just corrected this before publishing. But anyway, the definition is the same, and I can only say that I don't like it for the reason explained in the post. :) –  Alexander Shamov May 4 '13 at 22:15

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I think your counterexample is correct (and this also means that Eisenbud's Spec construction breaks down; by the way there is a global Spec construction which even works for locally ringed spaces, see here). The correct definition of a prime ideal sheaf is that the corresponding closed subscheme is integral. Equivalently, it should be irreducible and reduced. Whereas reducedness can be checked locally, this is not the case for irreduciblity. So here is a more down-to-earth description: $I \subseteq \mathcal{O}_S$ is prime if $I \neq \mathcal{O}_S$, for every open affine $U \subseteq S$ the ideal $\Gamma(U,I) \subseteq \Gamma(U,\mathcal{O}_S)$ is prime or $(1)$, and for every inclusion $V \subseteq U$ of opn affines the map $\Gamma(U,\mathcal{O}_S) / \Gamma(U,I) \to \Gamma(V,\mathcal{O}_S) / \Gamma(V,I)$ is injective.

PS: You may contact Eisenbud. He collects errata for his book.

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Thank you. In your down-to-earth description you are still only working locally on affines, so I believe my counterexample still applies here. And the last injectivity condition, I think, should be automatically true if $\Gamma(U,I)$ is prime and $\Gamma(V,I) \neq (1)$, and wrong if $\Gamma(V,I) = (1)$. Anyway, this is what I wanted to know, that there should be no issues besides this "global" irreducibility... –  Alexander Shamov May 4 '13 at 22:04
    
Hm, so what is the correct down-to-earth description? –  Martin Brandenburg May 5 '13 at 8:47

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