If $X\subseteq\mathbb{A}^N$ is a variety of dimension $n$, I would like to prove that $$\dim T_xX\geq\dim X=n$$ for any point $x\in X$.
My first attempt was to use the Krull's Principal Ideal Theorem, because if $\dim X=n$, then we need at least $N-n$ generators for $I(X)$, say $F_1,...,F_m$. Then, their differentials at $x$ define $T_xX$, since $T_xX=V(d_xF_1,...,d_xF_m)$, and by applying the above theorem to its coordinate ring, we conclude that $\dim T_xX\geq N-m$ and I can't continue, because $m\geq N-n$.
I came up with this question because the class notes I was reading define a point to be non-singular when $\dim X=N-\textrm{rank }J(x)$, where $J(x)$ denotes the Jacobian matrix at $x$, and claim that a point is singular if, and only if, $\textrm{rank }J(x)<N-n$ (why couldn't it be strictly greater instead?), from where we conclude that $\dim T_xX>\dim X$ (since $T_xX$ is the null space of $J(x)$ and we use the Rank-Nullity Theorem).