Let $V$ be an algebraic variety of dimension $m$ over an algebraically-closed field of characteristic $0$, and let $n<m$ and $U\subset V$ be a closed subset of $V$. Must there exist a subvariety $U\subset U'\subseteq V$ with $U'$ of dimension $n+1$? In the affine case, this is just asking if the solutions to a system of $m-n$ or more polynomial equations are all solutions to a system of $m-n-1$ equations giving a variety. It is foundational to algebraic geometry that a potentially reducible $U'$ exists. I can show the result using Lagrange interpolation for the projective case when $n=1$, but beyond that I am lost. I am not particularly learned in algebraic geometry.
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Maybe I am misinterpreting your question but isn't the union of the three coordinate axes in $\mathbb{C}^3$ a counterexample? – Derek Allums Sep 07 '23 at 08:40
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@DerekAllums As $V$ yes. Silly of me. I should have said $V$ is a variety. As $U$ I don't think so. They are all contained in a cone. Cones are irreducible. – Thomas Anton Sep 07 '23 at 11:37
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@Derek Allums. Why is this a counterexample? The union of the three coordinate axes in ${\bf C}^3$ is contained in the variety defined by the ideal $(z)\cap (x,y)$, which has dimension $2$. – Damian Rössler Sep 07 '23 at 12:51
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1@DamianRössler But this is reducible as the union of the $xy$-plane and the $z$-axis, no? – Derek Allums Sep 07 '23 at 13:38
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@Derek Allums. You are right - I overlooked that assumption, sorry. – Damian Rössler Sep 07 '23 at 21:04
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@DamianRössler Not a problem! Forced me to sanity check my claim which is always a good thing. – Derek Allums Sep 08 '23 at 08:12