# For any irreducible variety $X$ and any point $x$ in $X$, dim$\mathscr{T}(X)_x \geq$ dim$X$, with equality holds in a dense open subset of $X$

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.

1. Why is $K(Y)$ equal to $K(X)$?
2. Why are there any nonempty isomoarphic open sets in $X$ and $Y$? (Is there any birational morphism between $X$ and $Y$?)
3. 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.

-

Consider the homomorphism $K(X)\to K(Y)$ that takes $t_i$ to the function $T_i$ restricted to $Y$ (basically $t_i\mapsto t_i$). The homomorphism described is obviously onto, and since it isn't the zero homomorphism, it must be injective.

We can see that $f$ is irreducible, because if $f(T_0,\ldots,T_d)=G(T_0,\ldots,T_d)H(T_0,\ldots,T_d)$, then because of the minimality of $f$ as a polynomial of $T_0$, we can assume that $H(T_0,\ldots,T_d)$ only depends on $T_1,\ldots,T_d$. Since $f$ is monic, we must have that $H=1$.

For your second question, since $K(X)\simeq K(Y)$, we have that $X$ and $Y$ are birational, and thus have isomorphic open subsets.

For your third question, consider the tangent bundle $\Theta=\{(a,x)\in \mathbb{A}^s\times X:a\in\mathscr{T}_x\}$ (where $\mathscr{T}_x$ is the tangent space of $X$ at $x$). This is a closed set, since if $X$ is defined by equations $F_1=\cdots=F_m=0$, then $\Theta$ is defined by equations $d_xF_1(a)=\cdots d_xF_m(a)=0$, where $d_xF_i$ is the differential of $F_i$ at $x$. We have that the second projection $\pi:\Theta\to X$ is a regular map, and is also surjective. By dimension theory, we have that there exists a positive integer $a_X$ such that for every $x\in X$, $\dim\mathscr{T}_x=\dim \pi^{-1}(x)\geq a_X$, and the points $x$ where $\dim\mathscr{T}_x=a_X$ is a dense open subset of $X$ (see Shafarevich, "Basic Algebraic Geometry: Varieties in Projective Space", Theorem 7, page 76). The points that satisfy this equality are called nonsingular points of $X$.

To see that $a_X=\dim X$, we use that $X$ and $Y$ are birational. We have that the set of nonsingular points of $Y$ is open and dense (because of what we just said). It is easy to characterize the nonsingular points of a hypersurface in $\mathbb{A}^{n+1}$ (as is the case of $Y$): We have that $\mathscr{T}_{y,Y}$ is given by the equation $\sum_{i=1}^{n+1}\frac{\partial f}{\partial T_i}(y)(T_i-y_i)=0$ (where $y=(y_1,\ldots,y_{n+1})$). Let's assume that $y$ is nonsingular. Then $a_Y=\dim Y=n-1$ if and only if not all the $\partial f/\partial T_i$ are not identically 0 on $Y$ (since a variety defined by one equation either has dimension $n+1$ or $n$, in this case). If we're in characteristic zero, if all these partial derivatives are 0, we would have that $f$ is constant (which it clearly isn't). If we're in characteristic $p>0$, then $f=g^p$ for some polynomial, which isn't the case since $f$ is irreducible. Therefore the derivatives aren't identically zero, and so we have that $\dim\mathscr{T}_{Y,y}=a_Y=\dim Y=\dim X$ (since $X$ and $Y$ are birational).

Since $X$ and $Y$ have two isomorphic open subsets, we take $y$ nonsingular in the isomorphic subset of $Y$. We have that $\dim\mathscr{T}_{Y,y}=a_Y=\dim Y=\dim X$. Since the dimension of the tangent space is invariant under isomorphism, we have that the tangent space of the corresponding point in $X$ is equal to $\dim X$, and the theorem is proved.

Is that clear enough? Are there any details you still don't understand?

-
Thank you really very very much for the taking the time and making the effort to give me the over-30-line answer! I will read it carefully for every detail. Would you please allow me to ask a further question: is it true that any two nonempty open subsets of an algebric variety intersect nontrivially, i.e. every nonempty open subset of an algebraic variety is dense? I am thinking of this because I don't see the word "dense" in Theorem 7 you referred to. Thanks again sincerely. – ShinyaSakai Sep 27 '11 at 4:11
First of all, thanks for your kind words. About your question, if the variety $X$ is irreducible, then we have that $X$ is not the union of two closed sets. Now let's take $U_1,U_2\subseteq X$ open subsets. If $U_1\cap U_2=\varnothing$, we have that $X=(X\backslash U_1)\cup(X\backslash U_2)$, which contradicts the irreducibility of $X$. So yes, in an irreducible variety, any open set is dense. Any other questions you have, be free to ask! If anything's not clear in my answer, or if I made a mistake, just let me know. – rfauffar Sep 27 '11 at 11:23
I just can't express my gratitude. Please allow me to ask you the last two questions on this proof. Does "$X$, $Y$ are birational" means there is a birational morphism between the two? On the book, the definition of birational morphism is a morphism $\phi: X \rightarrow Y$ such that $\phi^*$ is an isomorphism of $K(Y)$ onto $K(X)$, but is the inverse true, i.e. does the isomorphism of function fields imply the existence of birational morphism? Best regards. – ShinyaSakai Sep 28 '11 at 4:08
Yes, to both of your questions. You can show that if $\psi:K(Y)\to K(X)$ is an isomorphism, then there is a function $\phi: X\to Y$ such that $\psi=\phi^*$. In the affine case, for example, if $\psi:k[Y]\to k[X]$ is a homomorphism and if $t_1,\ldots,t_m$ are the coordinate functions in the ambient space of $Y$ (let's say $\mathbb{A}^m$), then we take $\phi(x)$ to be $(\psi(t_1)(x),\ldots,\psi(t_m)(x))$. That's the general idea when we go to the quotient field (I'm leaving out a lot of details, however). – rfauffar Sep 29 '11 at 1:19
Thank you very much! Now I have learned a lot and understand why the proof proceeds like this because of your kind help. Thanks again. – ShinyaSakai Sep 29 '11 at 3:58