Assume that we're working in an algebraically closed field. Let $X \subsetneq \mathbb{A}^n$ be a Zariski-closed set. Is there a line in $\mathbb{A}^n$ that intersects $X$ in finitely many points? I need a hint.
Is there always a line that intersects a given Zariski-closed set in finite number of points?
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algebraic-geometry
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1Is $X$ supposed to be properly contained in $\mathbb{A}^n$? And should "algebraically closed set" be "algebraically closed field"? – 2012-12-29
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0@QiaochuYuan it was a typo :) About $X$, assume you're talking to someone who doesn't know the first thing about schemes. – 2012-12-29
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1@Alexei: I think Qiaochu's point is that if $X=\mathbb{A}^n$ the answer is trivially no (since any algebraically closed field is infinite). – 2012-12-29
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0@ZevChonoles Oh :) Yeah, I meant a proper subset. – 2012-12-29
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Suppose that a line $\ell$ intersects $X$ in infinitely many points. The intersection is an infinite Zariski closed subset of $X$ and $\ell$. Now $X$ may have lots of different infinite Zariski closed subsets (depending on what it is), but what about $\ell$?
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0Once again, AG is so easy once someone explained it :D Thanks! – 2012-12-29