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I am trying to solve Exercise 3.15 in Gathmann's 2014 notes. Given an irreducible affine variety $X$ such that $A(X)$ is UFD und $U$ like in the heading I want to show that the sheaf of regular functions on $U$ is already the whole coordinate ring $A(X)$.

From a previous exercise I know that regular functions on $U$ are given as a quotient of polynomials globally on $U$ if $A(X)$ is UFD i.e. $\varphi(x) = \frac{g(x)}{f(x)}$ for all $x \in U$ and $f,g \in A(X)$. It seems reasonable to show that $f$ is a unit using the fact that this equation kind of holds in dimension $2$ but I have no idea how to proceed at this point.

Felix G
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    This is algebraic Hartogs's Lemma. – Viktor Vaughn Mar 24 '20 at 15:51
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    For a solution phrased in terms of scheme: let $R = \mathcal{O}X(X)$ then $X=\text{Spec} R$, then $\mathcal{O}_X(U) = \cap{\mathfrak{p}\in U} R_\mathfrak{p}$, assumption on $U$ says it contains all prime ideals of height $1$, then use a result commutative algebra: for an integrally closed Noethrian ring (in particular, Noetherian UFD), intersection of localizations at height 1 prime ideals is the ring itself. – pisco Mar 24 '20 at 15:52
  • @pisco that looks like an answer to me - would you care to record it below? – KReiser Mar 24 '20 at 18:11

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This is a community wiki answer recording the discussion from the comments, in order that this question might be marked as answered (once this post is upvoted or accepted).

This is algebraic Hartogs's Lemma. – Richard D. James

For a solution phrased in terms of scheme: let $R=\mathcal{O}_X(X)$ then $X=\operatorname{Spec} R$, then $\mathcal{O}_X(U)=\cap_{\mathfrak{p}\in U} R_\mathfrak{p}$, assumption on $U$ says it contains all prime ideals of height 1, then use a result commutative algebra: for an integrally closed Noethrian ring (in particular, Noetherian UFD), intersection of localizations at height 1 prime ideals is the ring itself. – pisco

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