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Let $X$ be an irreducible variety. Then any open set $U\subset X$ is dense. Thus $U$ is irreducible as topological space. Consider $O_X(U)$ sheaf of regular functions on $U$.

If $U'\subset U$ is open in $U$, is $O_X(U)\to O_X(U')$ injective?

I think it is the case, if $f\in O_X(U)$ is sent to $0\in O_X(U')$, then $f|_{U'}=0$. So $U'\subset V(f)$ and $U=(U-U')\cup V(f)$. Clearly $V(f)=U$ by $U$ irreducible. Thus $f|_U=0$.

This is related to Extending regular function on normal variety from a subvariety of codimension 2. In this post, @MooS Answer says we know $\cap_{ht(p)=1}A_p=A$ implies restriction map injectivity. However this theorem requires normal variety. An irreducible variety is always normal?

user45765
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1 Answers1

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If $f\in O_X(U)$ in not zero, then it is not zero in some affine $V\subset U$ and and hence non-zero in the local ring at the generic point $\eta$.

This proves that the composition $O_X(U) \to O_X(U') \to O_{\eta}$ is injective, and so is the first map.

Chan Ki Fung
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