For any irreducible variety $X$ and any point $x$ in $X$, $\mathrm{dim}\mathscr{T}(X)_x \geq \mathrm{dim}X$, with equality holds in a dense open subset of $X$.
Here, $\mathscr{T}(X)_x$ denotes the tangent space of $X$ at $x$. Let $\mathscr{O}_x$ be the local ring of $x$, and $\mathscr{m}_x$ its maximal ideal, then $\mathscr{T}(X)_x$ is defined to be the dual vector space $(\mathscr{m}_x/\mathscr{m}_x^2)^*$ over$\mathscr{O}_x/\mathscr{m}_x$.
Part of the proof is as follows:
$K(X)$ is a separably generated extension of $K$, i.e., $K(X)$ is a separable algebraic extension of a subfield $L=K(t_1, \cdots, t_d)$, the latter being purely transcendental over $K$. The theorem of the primitive element allows us to find a single generator $t_0$ of the extension $K(X)/L$. Let $f(T_0) \in L[T_0]$ be its minimal polynomial. This defines a rational function $f(T_0, T_1, \cdots, T_d) \in K(T_0, T_1, \cdots, T_d)$, defined on an affine open subset of $\mathbb{A}^{n+1}$, where its set of zeros $Y$ is a hypersurface with function field $K(Y)$ isomorphic to $K(X)$. Some nonempty open sets in $X$ and $Y$ are isomorphic. The points $y \in Y$ where $\mathrm{dim} \mathscr{T}(Y)_y = \mathrm{dim}Y$ form a dense open subset of $Y$, so in particular $\mathrm{dim}\mathscr{T}(X)_x = \mathrm{dim}X$ for $x$ in some dense open subset of $X$.
There are three statements which I can't understand.
- Why is $K(Y)$ equal to $K(X)$?
- Why are there any nonempty isomoarphic open sets in $X$ and $Y$? (Is there any birational morphism between $X$ and $Y$?)
- Why do the points $y \in Y$ where $\mathrm{dim} \mathscr{T}(Y)_y = \mathrm{dim}Y$ form a dense open subset of $Y$?
This on page 40 of James E. Humphreys' Linear Algebraic Groups. Thanks very much.