Bound on dimension of tangent space of an affine variety

Question

I've been reading through my notes and the following fact is stated without any proof or justification: For an affine variety $X\subset\mathbb{A}^n$ and a point $p\in X$ we have $$dim_k T_pX\geq dim_p X$$ where $dim_p X$ denotes the local dimension of $X$ at $p$. At first I thought I could use the fact that $dim_k T_pX \geq n-N$ where $N$ is the size of a minimal generating set of the ideal defining $X$, plus Krull's height theorem but the more I think about it the more I convince myself that this is hopeless. Could someone point me towards the right direction?

Answer 1

You are right that this follows from Krull's principal ideal theorem. First note that $\dim_k T_pX$ is the minimal number of generators of the maximal ideal $\mathfrak m \subset \mathcal O_{X, p}$, because $T_p X = (\mathfrak m / \mathfrak m^2)^\vee$. Set $d = \dim \mathcal O_{X,p}$. Now by Krull's principal ideal theorem, any ideal $I \subset \mathcal O_{X,p}$ generated by $r$ elements has height at most $r$, so that $$\dim \mathcal O_{x,p} / I \geq d - r$$ Note that then even if $\operatorname{ht}(I) = r$ it could happen that $\dim \mathcal O_{X,p} / I > d - r$, if $X$ is not irreducible. Nevertheless this shows that $\mathfrak m$ cannot be generated by less then $d$ elements, because $\mathcal O_{X,p} / \mathfrak m$ is a field, so that $\dim \mathcal O_{X,p} / \mathfrak m = 0$.