three axis in $\mathbb{A}^3$ can't be defined by two functions

Asked Oct 20 '14 at 01:29

Active Oct 20 '14 at 01:41

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I am reading Shafarevich's book on Algebraic Geometry and in 1.6.5, exercise 3. He asks to prove that $X \subset \mathbb{A}^3$, which is the union of the three coordinate axis, can not be defined by two functions. i.e. $I(X)$ must be generated by at least three elements.

I am lost on how to solve this problem.

edited Oct 20 '14 at 01:41

Ben Grossmann

225,327

asked Oct 20 '14 at 01:29

Alex

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$I(X)=( x y, x z, y z )$ – orangeskid Oct 20 '14 at 02:33
See the hint for an analogous problem with another monomial ideal http://math.stackexchange.com/questions/976156/showing-that-an-ideal-is-not-generated-by-two-elements/ – orangeskid Oct 20 '14 at 02:35
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Two solutions: here http://math.stackexchange.com/a/92518/3217 and here http://math.stackexchange.com/a/510997/3217 – Georges Elencwajg Oct 20 '14 at 07:59
@GeorgesElencwajg: Thanks! Those two answers were really helpful! – Alex Oct 21 '14 at 03:10
See also http://math.stackexchange.com/questions/92505/determining-the-generators-of-ix – Jo Wehler Nov 05 '14 at 04:57

three axis in $\mathbb{A}^3$ can't be defined by two functions

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