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Having browsed through some books and several online discussions (here and on MathOverflow) on algebraic geometry, it seems to me that the following is true:

Suppose $\mathbb K=\mathbb R$ or $\mathbb K=\mathbb C$. Let $f\colon\mathbb K^n\to\mathbb K^k$ be a polynomial map (each component a polynomial) and let $V\subset\mathbb K^n$ be its vanishing set. Fix a natural number $r$ and suppose that on $V$ the rank of the Jacobian $Df$ never exceeds $r$ on $V$. If the rank of $Df$ is exactly $r$ at a point $p\in V$, then in a (Euclidean or Zariski) neighborhood of $p$ the variety $V$ is a smooth submanifold of dimension $n-r$.

Can someone provide a reference?

Remarks:

  • Many seem to point to a claim like this being true (e.g. Hartshorne in the opening of §I.5 of his book), but I have been unable to track down this statement as a citable theorem.
  • I would like to be able to cite this in a context where the audience knows nothing at all about algebraic geometry. I want to be able to verify the assumptions for any concrete $f$.
  • This is easy to prove using the implicit function theorem when $r=k$, but otherwise I would not consider it obvious to my audience.
  • Small variations of the claim are welcome; I don't expect to find this literally. A theorem saying "near smooth point of a variety the variety is in fact a smooth manifold" is fine if the definition of a smooth variety agrees with mine or is otherwise a similarly easy property of $f$ to check.
  • I'm not looking for a proof but a reference. My previous question was about understanding a feature of the proof.
Joonas Ilmavirta
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  • This doesn't seem quite right: Take $n = k = 1$ and $f(x) = x^3$ which vanishes only at the origin where it also has a critical point. So we may take $r = 0$, but $V$ is a smooth manifold of dimension $0 \neq 1 - 0 = 1$. Perhaps that should be codimension? – Ben Steffan Jun 02 '25 at 16:56
  • @BenSteffan It seems I do indeed need something to rule that behavior out. Maybe that the ideal generated by $f$ is radical? This is easy enough to check with computer algebra software. – Joonas Ilmavirta Jun 02 '25 at 17:11

1 Answers1

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This is a consequence of the implicit function theorem, so you can just call it that. A variant of this is sometimes called the constant rank theorem.

For $\mathbb K = \mathbb C$, you need a holomorphic implicit function theorem, which exists (see e.g. Griffiths and Harris Foundations of Algebraic Geometry).

red_trumpet
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  • Thanks, the constant rank theorem looks very useful. Do you have a reference for that other than Wikipedia? It only makes the statement without citing anything. – Joonas Ilmavirta Jun 03 '25 at 02:39
  • @JoonasIlmavirta That is treated in Bröcker and Jänich's Introduction to Differential Topology. – red_trumpet Jun 03 '25 at 05:14