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We work over the complex numbers, $\mathbb{C}$. Let $X$ be a smooth affine variety. Let $U$ be an open subvariety of $X$. Then we have a natural map $$ i:\operatorname{Spec}\mathbb{C}[U]\to X $$

My question is: is this map an open immersion?

Note: The exact same question was asked here, except no smoothness assumption was made.

hm2020
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freeRmodule
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  • There is an exercise in Hartshorne (Ex.II.2.16) saying that if $A:=\Gamma(U, \mathcal{O}U)$ and $U$ is quasi compact with a finite open cover of affines, there is for any element $f\in A$ an isomorphism $\Gamma(U_f,\mathcal{O}{U_f}) \cong A_f$. Maybe this implies that the map is an open embedding with these hypotheses. – hm2020 Sep 03 '21 at 15:20
  • Let $X:=Spec(B)$. Since $U$ is open, there is an open cover $\cup_i D(f_i) =U$ with $f_i \in B$. There is a canonical restriction map $B \rightarrow A$ mapping $f_i$ to elements $g_i \in A$. If somehow $D(g_i):=Spec(A_{g_i})$ is an open cover of $U$, such that the restriction of your map $i$ to $D(g_i)$ is the inclusion $D(f_i) \subseteq X$. – hm2020 Sep 03 '21 at 15:26
  • @hm2020 I agree with what you write, but the problem is the difference between $\operatorname{Spec}\mathbb{C}[U]$ and $U$ itself. Your proof only shows that $U\to X$ is an open embedding, which we know by assumption. Keep in mind my statement is not true in complete generality; for a counterexample in a non-smooth case, see the cited question. – freeRmodule Sep 03 '21 at 15:50

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I think the answer is positive and the following is a proof. All we will assume is that $X$ is locally factorial, which is implied by smoothness. I would really appreciate it if someone can check it though!

Let $Z$ be the complement of $U$ in $X$, as a topological space. Let $\xi_1,\dots,\xi_k$ be the codimension 1 primes coming from $Z$. Each $\xi_i$ defines a pure codimension-1 subvariety of $X$, call it $Y_i$, and all of them are contained in $Z$. Since $Y:=Y_1\cup\cdots\cup Y_k$ is pure codimension 1, and $X$ is locally factorial, $Y$ is an effective Cartier divisor. Thus the open set $U':=X\setminus Y$ is affine.

Now the complement of $U$ in $U'$ is codimension 2, so by Hartog's lemma we have an isomorphism $$ \mathbb{C}[U']\to\mathbb{C}[U] $$ coming from restriction of functions. It follows that $U'=\operatorname{Spec}\mathbb{C}[U]$, and we are done.

freeRmodule
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