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Suppose $X$ is an irreducible scheme, $U \subset V$ open subsets of $X$, does it hold that $\rho_U^V:O(V)\to O(U)$ injective? Generally under what conditions does it hold?

Actually it is related to an exercise in Liu Qing's book p67,Ex4.11:

Let $f:X\to Y $ be a morphism of irreducible schemes, show that the following are equivalent:

(2)$f^{\#}:O_Y \to f_*O_X$ is injective

(3)for every open subset $V$ of $Y$ and every open subset $U\subset f^{-1}(V)$, the map is injective.

to deduce (3) from (2) I had hoped the restriction map should be injective.

But now I don't know how to deal with it..

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Isn't $\rho^V_U$ going in the opposite direction? – a.r. Nov 9 '13 at 11:24
I edited it, thanks – mqx Nov 9 '13 at 13:08
The morphism in (2) doesn't make sense. – Georges Elencwajg Nov 9 '13 at 13:29
Sorry I corrected it, thanks – mqx Nov 9 '13 at 13:31
In the book, $X, Y$ are integral ! – Cantlog Nov 9 '13 at 18:12
up vote 5 down vote accepted

No, the restrictions needn't be injective, even for affine schemes. Here is a counterexample:

Let $k$ be a field, $A$ the ring $A=k[X,Y]/(Y^2,XY)=k[x,y]$ and $S=\operatorname {Spec} (A)$ the corresponding affine scheme.
The restriction morphism from $U=S$ to $V=D(x)$$$\rho:\mathcal O(S)=A \to \mathcal O (D(x))=A_x$$ is not injective because it sends $y\neq 0\in \mathcal O(S)=A$ to $y|D(x)=\frac {y}{1}=0\in \mathcal O (D(x))=A_x$ .
[Why is $\frac {y}{1}=0\in A_x$ ? Because $xy=0$ and $x$ is invertible in $A_x$]

The existence of the nontrivial nilpotent $y$ in this example is not coincidental: in a scheme that is irreducible and reduced (such schemes are called integral) all restriction maps between open subsets are indeed injective.

In answer to a request of the OP in his comment below, the quickest way to prove that the restriction map $\rho_U^V:O(V)\to O(U)$ is injective in the integral case is to compose it with the canonical morphism $\mathcal O(U)\to O_{X,\xi}$ into the generic stalk and to remark that the composition $\mathcal O(V)\to O_{X,\xi}$ is injective as a consequence of Qing Liu's Proposition 4.18 (b), page 65.
[Some kid on the block will remind you that $v\circ u$ injective $\implies u$ injective]

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If X is integral, $s \in \Gamma(V)$ is mapped to $0\in \Gamma(U)$ How do we get a contradiction? Could you explain your last claim a little bit? Thanks! – mqx Nov 9 '13 at 13:29
Dear mqx: since my example is not integral you can't use the notation in it to ask questions about integral schemes! The best I can say is that the corresponding reduced scheme $S_{red}=\operatorname {Spec} (A_{red})$ with $A_{red}=k[X,Y]/(Y,XY)=k[X,Y]/(Y)=k[x,y]$ has $y=0\in \mathcal O_{S_{red}}(S_{red})=A_{red}$ and the obstruction mentioned in my post to the restriction being injective no longer applies . – Georges Elencwajg Nov 9 '13 at 13:38
I am sorry that I used unclear notations..could you help me with the case when X is an integral scheme, to show the restriction map is injective? – mqx Nov 9 '13 at 13:48
Dear mqx, your notation was not unclear: I was finishing editing my answer while you wrote your second comment! I hope the edit answers your question. – Georges Elencwajg Nov 9 '13 at 14:00
Does the equivalence in exercise still hold for just irreducible case? – mqx Nov 9 '13 at 14:04

Think of the affine case. So let $X= Spec\, A$ for some ring $A$. Then the basic open sets are the sets of the form $D_f = Spec \, A_f$. The restriction map $O(X) \to O(D_f)$ is precisely the localization map $A \to A_f$. This is injective precisely when $f$ is a non-zero divisor.

This argument could be globalized to the case when $X$ is a scheme. So at least when $X$ is an integral scheme, the restriction maps should in general be injective.

Edit: Of course, as Zhen Lin points out, if $U=\emptyset$, then the restriction map is just the zero map $A \to 0$, so it is not injective in that case.

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Except if $U = \emptyset$... – Zhen Lin Nov 9 '13 at 11:28

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