User li zhan - MathOverflow

Standard exact sequence for a complete intersection

2012-08-05T02:55:05Z

Suppose $P$ is a variety over $\mathbb{C}$(in my case $P$ might have slight singularities), and $V=\cap_{i}^{r} D_i$ is a complete intersection. Then, it is claimed exists the so called "Standard exact sequence for a complete intersection":

$$0 \to O_{V}(-D_1) \oplus \cdots \oplus O_{V}(-D_r) \to \Omega_{P}^{1}|_V \to \Omega_{V}^1 \to 0.$$

Does anyone know how to show the above exact sequence? Or point out some reference having discussiong about such things? Moreover, what are the corresponding sequence for the $p$-forms (i.e. the sequence tells something about $\Omega_{P}^{p}|_V $ and $\Omega_{V}^p $ )?

Vanishing of higher cohomology

2012-05-13T04:40:33Z

If $M$ is a manifold of dimension $n$, does singular cohomology $H^i(M, \mathbb{C})$ vanish when $i > n$ ?

If $M$ is an algebraic variety over $\mathbb{C}$, equipped with ordinary topology, can one say something about the vanishing of higer singular cohomology?

A characterisation of tame ramification

2011-06-07T04:13:21Z

The following is the statement from Algebraic Number Theory by Neukirch (Chapter 2 Proposition (7.7) p155)

Suppose $K$ is Henselian field, $p=char(\kappa)$, the character of the residue field of $K$. A finite extension $L\K$ is tamely ramified if and only if the extension $L/T$, ($T$ is the maximal unramified subextension of $L/K$) is generated by radicals $\quad L=T( \sqrt[m_1]{a_1},\dots, \sqrt[m_r]{a_r})$, such that $gcd(m_i, p)=1$.

For "$\Rightarrow$" direction, the proof given in the book is correct, but it should be pointed out that $"a_i"$s come from $T$.

The proof of "$\Leftarrow$" direction is highly suspect. First of all, what's the right statement? There are at least two ways:

(1) $K$ is a Henselian field, for $a_i \in K,$Let$ L=K( \sqrt[m_1]{a_1},\dots, \sqrt[m_r]{a_r}), \quad gcd(m_i,p)=1, \quad p=char(\kappa)$. Then $L/K$ is a tamely ramified extension.

(2) Same as (1) + $K$ is just the maximal unramified subextension of $L/K$ (i.e. $L/K$ is totally ramified ).

Does anyone know the proof of either statement? In addition, if $L/K$ happens to be a finite Galois extension (or maybe you only need simple extension), is it true $L=K(\sqrt[m]{a})$ form?

Does totally ramified extension really exist?

2011-06-06T08:22:23Z

The answer is certainly "Yes", but this is the problem I met in Algebraic Number Theory by Neukirch. I guess that I must be doing something wrong, since otherwise I will get the statement "There are no totally ramified extensions except the trival ones".

Let $K$ be Henselian field, $L/K$ be a finite, totally ramified extension. Let $\lambda$ and $\kappa$ be the residue field of $L$ and $K$ respectively. Because $L/K$ is totally ramified, $K$ is the maximal unramified subextension, so we have $\lambda=\kappa$. If $L\ne K$, let $a \in L-K$. Since the valuation is non-trival, by multiplying by an element in $K$, we can suppose $a \in O_L$, the valuation ring of $L$.

Because "the valuation ring of $L$ is the integral closure of the valuation ring of $K$ in $L$ "(P144, Chapt 2 Theorem (6.2) of Neukirch), let $f(x) \in O_K[x]$ be the minimal polynomial of $a$ in $O_K$, where $O_K$ is the valuation ring of $K$. (One can prove $f(x)$ is monic, and I guess it may differ from the minimal polynomial over $K$.) Let $\bar{f}(x)$ be the corresponding polynomial over the residue field $\kappa$. It must be the minimal polynomial of $\bar{a} \in \lambda$, because otherwise, by Hensel's lemma, $\bar{f}$ admits a factorization in $\kappa[x]$ implies $f$ admits a factorization in $O_K[x]$. But since $\lambda=\kappa$, we get $deg(\bar{f})=deg(f)=1$ ($f$ is monic). This means $a\in K$, a contradiction. This means $L=K$ !

An element in the product of schemes

2011-06-01T03:44:51Z

Let $ f : X \to S, g : Y \to S$ be the scheme morphisms, and $ X \times_S Y$ be the product of shemes. Let $ z \in X \times_S Y$ , and $ x=p(z) , y=q(z) ,$ where $ p: X \times_S Y \to X, q: X \times_S Y\to Y $ are projections. Suppose $ s=f(x)=g(y) \in S$ and $x\in U \subset X, y\in V \subset Y, s\in W \subset S, f(U) \subset W, g(V) \subset W$ where $U, V, W$ are arbitrary open sets. Is it ture $ z\in U \times_W V$ ?

(The question should be read as : Is it ture that $ z$ is in the image of open immersion $U \times_W V \to X \times_S Y$ )

Just a little comment: this can be used to prove monomorphism (particularly open and closed immersions) is seperated. The reason is, by the above result, one can pick up an open cover of $ X \times_Y X$ as $ U \times_V U$, because of the monomorphism property, two projections are the same, so one can choose the same open set $U$, and by the morphism of affines is seperated, we are done.

Comparation of dimensions of rings

2011-05-18T08:12:27Z

Let $k$ be a field, $A$ be an integral domain, $B \subset A$, and $A, B$ are both finitely generated $k$ algebra. Let $p \subset B$ be a prime ideal. Suppose there exists prime ideals $q \subset A$, such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion. Then, is it true $ dim A_q \leq dim B_p$ ?

For the geometric meaning, it comes from the exercise of Chapter 2, 3.22(a) of Hartshorne, where: Let $ f: Spec(A) \to Spec(B) $ be a dominant morphism, $p \in Spec(B), Y'=$ {$ \bar{ p }$} (the closure of {$p$}) and $Z$ be an irreducible component of $f^{-1}(Y')$, whose generic point $q$ maps to $p$, then show that $ codim(Z,X) \leq codim(Y',Y)$.

I guess, everything translates faithfully to the above algebra fact except "$f$ dominant " is weakend by " $ B \to A$ is injective".

Comment by Li Zhan

2012-08-07T23:36:53Z

@Will Sawin I find the comments for my first question is helpful, however, nothing has been said about the second one, i.e. the exact sequence about p-form. I will be very appreciate if you can say something in details.

Comment by Li Zhan

2012-08-06T00:04:03Z

@Mariano Suarez-Alvarez I didn't realize you had done the editing! I just keep doing many times. Sorry about that!

Comment by Li Zhan

2012-08-06T00:02:47Z

@Ryan Reich Thank you for your editing!

Comment by Li Zhan

2012-08-05T03:15:03Z

Could do say something more about your answer? What do you mean by adjunction exact sequence?

Comment by Li Zhan

2012-08-05T03:10:09Z

Sorry about the appearance of my post, but why it can code through in mathstackexchange.com but not mathoverflow.com (which I think it might more suitable to put it here)?

Comment by Li Zhan

2012-05-30T05:13:53Z

@Donu Arapura Is this a counterexample of "4) If X is paracompact, Cech cohomology coincides with Grothendieck cohomology for ALL SHEAVES" posted by the answer of Georges Elencwajg?

Comment by Li Zhan

2012-05-14T09:27:13Z

@Charles Matthews Thank you, I would like to accept this as a partial answer to my question.

Comment by Li Zhan

2012-05-14T09:21:12Z

@Zhen Lin, I agree, and obiously it was me post the same question. But NO satisfied answer has been got. That's why I post it here, where more attention will be drawn by experts.

Comment by Li Zhan

2011-06-07T11:59:38Z

Hi Alex Bartel, I cannot see the equivalence of two statement. As you said,$L$ is tamely ramified over the maximal unramified intermediate extension. But how do you know this maximal unramified subextension is just $K$?

Comment by Li Zhan

2011-06-06T09:58:40Z

Great! Thank you very much!

Comment by Li Zhan

2011-06-02T03:24:08Z

Thank you for your detailed explanation! I did not know that : In every category a monomorphism has the property that the diagonal is an isomorphism. Could you please say more about this? I know it boil down to prove the affine case (in the category of scheme), but I cannot prove this. As for the open immersion $U \times_W V$, it comes from the construction of product : the product is the gluing of shemes of the form $U \times_W V$, and by the gluing lemma, one knows each piece has an open immersion to the target scheme. – Li Zhan 0 secs ago

Comment by Li Zhan

2011-06-02T03:02:42Z

Nice argument! It is essentially like the second approach in the second answer, but successfully avoid complicated argument.

Comment by Li Zhan

2011-05-19T09:30:34Z

Dear Georges Elencwajg, You are right! Thank you! There are two commnets: 1. I do mean $q $ can be any minimal prime satisfies that condition, but I guess, it is not hard to prove actually this is the only prime by the uniqueness of generic point of integral scheme. 2. I don't quite understand the ring $A_q \otimes_B k(p)$ is the local ring of the generic point of fiber. But for me, I can prove this ring is isomorphism to the ring $A_q / pA_q$, and the by the minimality of $q$, its dim is zero.

Comment by Li Zhan

2011-05-19T03:42:13Z

Dear Konstantin Ardakov，could you say more about how to apply Noether Normalization to $B$ to get $p$ generated by $dim(B_p)$ elements? After konwing this, it will be perfect for applying Krull's theorem, because taking into account the minimality of $q$ , one can prove prime ideal $qA_q$ is just minimal over $pA_q$ .

Comment by Li Zhan

2011-05-19T03:03:24Z

Dear Karl Schwede, thank you! The direction you point out used to make me unease when translate it into algebra facts.