Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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?

share|improve this answer
    
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. –  Robert Auffarth 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). –  Robert Auffarth 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.