Let $X \subseteq \mathbb{A}^n$ be algebraic, and let $x$, $y \in \mathbb{A}^n - X$. How do I see that there exists $f \in I(X)$ with $f(x) \neq 0$ and $f(y) \neq 0$.
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3$\begingroup$ Choose $g\in I(X)$ such that $g(x)\neq 0$, since if $I(X)(x)=0$, $x\in X$, and likewise $f(y)\neq 0$. Let $x$ and $y$ differ in the first coordinate without loss of generality, by application of a rotation. Then the function $(x-x_1)g(x)+(x-y_1)f(x)\in I(X)$ then has the desired properties. $\endgroup$– PaxCommented Sep 28, 2015 at 21:16
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Given an affine scheme $S=\operatorname {Spec (R)} $ and a closed subscheme $T=V(J)\subset S$, the restriction mapping $$\mathcal O(S)=R\to \mathcal O(T)=R/J$$ is obviously surjective.
Applying this to $S=\mathbb A^n, T=X\cup \{x,y\}$ and taking for $f_0\in \mathcal O(T)$ the function equal to $0$ on $X$ and $1$ on $x$ and $y$, we see that we can extend $f_0$ to a regular function $f\in \mathcal O(\mathbb A^n)$, which will thus satisfy the requested conditions.
Remark
The trivial but useful surjectivity result in the first sentence is the algebraic geometers' version of the Tietze extension theorem, an analogy sadly not mentioned in algebraic geometry books.
answered Sep 28, 2015 at 22:17
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$\begingroup$ @ Elencwajg - are you sure Hartshorne doesn't mention it. I have a dim memory ( in general and of this fact) $\endgroup$– mehCommented Sep 28, 2015 at 23:26
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$\begingroup$ Dear @agensky, let me be specific: I claim that neither Hartshorne nor any other algebraic geometry book alludes to Tietze's extension theorem in this context . Users are welcome to give a reference contradicting this claim :-) There have been several analogous interpolation questions on Mathematics StackExchange and they are always answered in an explicit ad hoc way, just as Chanler did in a comment to this question. I answered this question in order to show the power of this underappreciated extension result, which for example also yields a geometric proof of the Chinese Remainder Theorem. $\endgroup$ Commented Sep 29, 2015 at 6:48