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.

Let $Y$ be a scheme of finite type over an algebraically closed field $k$. Show that the function $\phi(y) = dim_k(m_y/{m_y}^2)$ is upper semicontinuous on the set of closed point of $Y$ (i.e. for any point $y$, there exists an open neighborhood $U$, such that for any $x \in U, \phi(y) \geq \phi(x)$ ).

I have two thoughts about this problem:

1) If $y$ is a smooth point, then using the property that singular set is closed, one can show semicontinuity. So the difficulty comes from the singular point.

2) I would like to using semicontinuity theorem of cohomology of fibers, but I don't know how to construct the corresponding coherent sheaf.

share|improve this question

2 Answers 2

up vote 3 down vote accepted

Since the question is local you may assume that $Y=Spec(A)\subset \mathbb A^N_k $, where $A=k[X_1,...,X_N]/(f_1,...,f_m)$.
In other words $Y$ is the fiber of $0$ for the morphism $f=(f_1,...,f_m):\mathbb A^N_k \to \mathbb A^m_k$ .
As in good old calculus we have a Jacobian matrix with value for each closed $x\in Y$ : $$J(x)= (Jac (f))(x) =(\frac{\partial f_i }{\partial x_j}(x)) \quad (i=1,...,m \;; j=1,...,N) $$
The number you are interested in is exactly the nullity of that matrix:
$$ \phi(x)=dim(T_x(Y))=dim_k( ker\:J(x))=m-rank (J(x)) $$
The conclusion follows : if $\phi(y)=d$, then some $(m-d)\times (m-d)$ minor of $J(y)$ is $\neq 0$.
It will remain $\neq0$ for all $x$ in a neighbourhood of $y$ and thus in that neighbourhood we will
have $rank (J(x))\geq m-d$ so that $\phi(x)=m-rank (J(x))\leq d$ as desired.

Note that neither (non)-singularity nor cohomology are evoked.

share|improve this answer
    
As Georges shows, this is classical algebraic geometry. I wonder whether a more general statement is true: if $X$ is a locally noetherian scheme, then is the function $x \mapsto \dim_{k(x)} \mathfrak{m}_x / \mathfrak{m}_x^2$ upper-semicontinuous? –  Andrea Apr 23 '12 at 18:25
2  
@Andrea: consider the affine line over $\mathbb C$. –  user18119 Apr 24 '12 at 1:12
    
@Georges Elencwajg Thank you so much for the proof!! –  Li Zhan Apr 24 '12 at 2:10
    
@Andrea QiL is right, since $Spec(\mathbb{C}[x])$ is smooth, $\phi(p)=dim(\mathbb{C}[x]_p)=1\quad $ except $p=0$, at that case, $\phi(0)=0$. But thank you all the same for making the clarification for the result. –  Li Zhan Apr 24 '12 at 2:14
    
I thank QiL for his usual clarity. My question was really stupid. –  Andrea Apr 24 '12 at 16:08

I think you can apply semicontinuity to sheaf of differentials.

But I don't see why affine line over C is a counterexample to andreas's question.

share|improve this answer

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.