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Let $Y\subseteq \mathbb{A}^n$ be algebraic. Assume $Y$ is irreducible.

Put $\mathfrak{p}=I(Y)$ , the ideal of $Y$ in $k[x_1,...,x_n]$.

Is it true that $height (\mathfrak{p})+ dim k[x_1,...,x_n]/\mathfrak{p} = \dim k[x_1,...,x_n]$?

This seems to have been used in a proof in lectures. Though I am not sure how to prove it?

Clearly $\mathfrak{p}$ is prime.

Could some care to elaborate on the intuition behind it please?

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    This should be in the literature somewhere, definitely in EGA. Consult Thm 5.6.8 of these notes. Note that polynomial rings are affine $k$-algebras and "coht" is essentially synonym to the dimension of the quotient ring. Intuition: dimension measures "#independent parameters". By Krull's height theorem the height also measures #independent parameters" needed to define the subspace for the prime ideal. If you subtract these two numbers you should be left with #independent parameters" of the subspace, i.e. the dimension. Rough idea.. – Thomas Preu Aug 06 '22 at 20:03
  • This is theorem 11.2.9 in Vakil's Foundations of Algebraic Geometry, available at http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf , but the proof is given as a series of exercises. – David Lui Aug 06 '22 at 20:08
  • @ThomasPreu no, that's not correct. See here for more details. – KReiser Aug 06 '22 at 20:39
  • My bad. Will remove the comment. – Thomas Preu Aug 06 '22 at 20:46
  • @KReiser What exactly is incorrect? Is the proof given in the linked notes incorrect? Or is it the independent parameters part that is incorrect? – David Lui Aug 06 '22 at 21:27
  • @DavidLui There was another (now deleted) comment claiming the result in the question is a consequence of a ring being (universally) catenary. Unfortunately that's not correct, as the linked answer shows. I did not mean to say anything about the first comment. – KReiser Aug 06 '22 at 21:43

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