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I've been told that given a polynomial map $f:X\to Y$ in characteristic zero, there exists an open dense subset $U$ of $X$ such that for all points $x$ in $U$, the rank of the derivative of $f$ in $x$ equals the dimension of the image of $f$. Could someone please explain to me why this is true, or direct me to some reference where this is explained?

Jasmine
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    A reference is Shafarevich, Basic Algebraic Geometry Volume 1, II.6.2 Theorem 2. (To apply the statement as written in your case, you need to remove the singular locus of $X$, but that doesn't cause any problems.) –  Jun 20 '14 at 16:05
  • Could someone elaborate on the reference given in the previous comment? The theorem in question seems much more general and I'm having trouble applying it to the specific case in the question. – Dániel G. Jan 03 '19 at 22:22

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