most recent 30 from http://mathoverflow.net 2013-06-18T22:01:25Z http://mathoverflow.net/feeds/user/10408 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection evgeniamerkulova 2010-12-13T13:50:27Z 2013-03-06T15:54:43Z

<p>Consider a ring $A$ and an affine scheme $X=SpecA$ . Given two ideals $I$ and $J$ and their associated subschemes $V(I)$ and $V(J)$, we know that the intersection $I\cap J$ corresponds to the union $V(I\cap J)=V(I)\cup V(J)$. But a product $I.J$ gives a new subscheme $V(I.J)$ which has same support as the union but can be bigger in an infinitesimal sense. For example if $I=J$ you get a scheme $V(I^2)$ which is equal to "double" $V(I)$.</p> <p>Vague Question : What is geometric interpretation of $V(I.J)$ in general?</p> <p>Precise question : When is $I\cap J=I.J$? Everybody knows the case $I+J=A$ but this is absolutely not necessary. For example if $A$ is UFD and $f,g$ are relatively prime then $(f).(g)=(f)\cap(g) $ but in general $(f)+(g)\neq A$ (e.g. $f=X, g=Y \in k[X, Y]$)</p> <p>Thank you very much. </p>

http://mathoverflow.net/questions/91106/local-rings-of-non-closed-points/91111#91111 evgeniamerkulova 2012-03-13T20:15:43Z 2012-03-13T20:15:43Z

<p>Since problem is local assume scheme is $Spec(R)$.Nonclosed point is prime ideal $P$ and can find maximal ideal $M \supset P$. Then $R_P=(R_M)_{PR_M}$ and use Serre theorem and hypothese that $R_M$ is regular.</p>

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective evgeniamerkulova 2011-02-13T18:25:35Z 2011-02-13T21:46:26Z

<p>Take scheme morphism $f: X\to Y$ and suppose $f$ surjective. If $y \in Y$ can one find affine open $V \subset Y$ containing $y$ and affine open $U \subset X$ such $f(U) = V$ ? Thank you.</p> <p>Later: Very good answer of Kevin shows it is not true. Is there hypothese which make it true ? For example $X$ irreducible and/or $f$ faithfuly flat ?</p>

http://mathoverflow.net/questions/44005/intuition-for-rational-functions evgeniamerkulova 2010-10-28T19:32:07Z 2010-11-02T22:44:28Z

<p>I asked this on mathematics stack exchange and did not receive answer . I hope it is good manners to ask here. Thank you very much.</p> <p>Let $X$ be integral scheme and $\mathcal K$ sheaf of rationnal functions on $X$. For any point $y\in X$ different of generic point we know that fiber of $\mathcal K$ (defined as usual as $\mathcal K _y / \mathcal m_y \mathcal K_y$) is zero. I'll be very gratefull if you explain intuitively why this is so, in language of restriction of $\mathcal K$ to reduced subscheme $Y=\overline{\{y \} }$. I have difficulty because many rationnal functions on $X$ can be restricted to nonzero rationnal functions on $Y$ . How is that compatible with fiber of $\mathcal K$ equals zero at $y$? </p>

http://mathoverflow.net/questions/58947/is-a-non-compact-riemann-surface-an-open-subset-of-a-compact-one/58948#58948 evgeniamerkulova 2011-03-20T08:15:36Z 2011-03-20T08:15:36Z

I do not understand. what is genus of $\mathbb C \setminus K$ for $K$ is Cantor set in $[0,1]$ ?

http://mathoverflow.net/questions/56415/projective-dimension evgeniamerkulova 2011-02-23T18:58:12Z 2011-02-23T18:58:12Z

@Fernando Muro: Obviously question is not obvious to asker: how is your comment helping? You come here for helping, no ?

http://mathoverflow.net/questions/55312/problem-in-rick-miranda-finding-genus-of-a-projective-curve evgeniamerkulova 2011-02-14T18:19:19Z 2011-02-14T18:19:19Z

roy, you are right that cohomology was probably asked and very well answered.I don't know how is called line + cubic curve. And you remind me very nice experience: when I learned english, all Americans were named Smith. I am very happy to meet real one: how do you do Mr. Smith?

http://mathoverflow.net/questions/55312/problem-in-rick-miranda-finding-genus-of-a-projective-curve/55317#55317 evgeniamerkulova 2011-02-14T13:59:47Z 2011-02-14T13:59:47Z

Cohomology computation is correct but proves not that intersection is elliptic curve: intersection could be two rational curves, line + cubic, if quadrics have common line. But does not happen with explicit equations here.

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55337#55337 evgeniamerkulova 2011-02-14T12:36:57Z 2011-02-14T12:36:57Z

Mr. Ventullo I am very sorry that cannot accept two answers and that I can give you only one vote: your answer merits more !

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55346#55346 evgeniamerkulova 2011-02-14T12:33:05Z 2011-02-14T12:33:05Z

Professor Liu thank you for answer, but I am not sure I understand it correctly because a word lacks between &quot;don't&quot; and &quot;openess&quot;! For local morphism of local rings flat= faithfully flat, right?

http://mathoverflow.net/questions/55312/problem-in-rick-miranda-finding-genus-of-a-projective-curve evgeniamerkulova 2011-02-14T12:14:28Z 2011-02-14T12:14:28Z

Dear Matt Emerton, I am not &quot;demanding more&quot;: I think such word is very strong in english and I would not use it. Also Technically, as of course you know, the question is more complicated because two quadrics could have common straight line and then intersection is reducible and not elliptic curve (it is not case here). Mr.Voloch's reference says skew elliptic curve must be intersection of two quadrics but does not speak of reverse. And you gave yourself wonderfull answer to beginner question : &quot;why nilpotents in algebraic geometry&quot;. Please continue give us such great answers in future.

http://mathoverflow.net/questions/55312/problem-in-rick-miranda-finding-genus-of-a-projective-curve evgeniamerkulova 2011-02-14T01:07:25Z 2011-02-14T01:07:25Z

Thank you for given your opinion , Mr.Voloch.

http://mathoverflow.net/questions/55312/problem-in-rick-miranda-finding-genus-of-a-projective-curve evgeniamerkulova 2011-02-14T00:52:14Z 2011-02-14T00:52:14Z

Mr. Voloch: if &quot;other books&quot; explain well, could you give title and page please? If not how can you be certain ? Can other people who wants to close please give solution before closing: certainly it will be to them very easy.

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55346#55346 evgeniamerkulova 2011-02-13T22:43:09Z 2011-02-13T22:43:09Z

So now we deduce thanks to your wonderfull answer following result, because we can reduce to affine case. If $f:X \to Y$ is faithfully flat morphism of integral schemes and $X$ is normal, then also $Y$ is normal.( Maybe should add $f$ locally of finite presentation to be sure $f$ is open?). Is that true, Professor?

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55346#55346 evgeniamerkulova 2011-02-13T22:29:59Z 2011-02-13T22:29:59Z

Wonderfull, Professor! This is very amusing because source of question is your book! In example 3.5 of 4.3.1 you say that normalization of integral scheme is flat only if scheme was already normal but you do not give proof and say to do exercise 1.2.10 (which is affine case) . I can do exercise just with faithfully flat assumption: if $A\subset B$ is faithfully flat ring extension and $B$ is integrally closed domain, then $A$ is also integrally closed ( no finiteness or noether assumption and I do not assume that $B$ is integral closure of $A$). Is that correct?

http://mathoverflow.net/questions/55333/surjective-implies-local-affine-surjective/55337#55337 evgeniamerkulova 2011-02-13T19:39:43Z 2011-02-13T19:39:43Z

Thank you very much: this is perfect answer!

http://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection/49262#49262 evgeniamerkulova 2010-12-13T14:36:56Z 2010-12-13T14:36:56Z

I think this is just formal translation of multiplication langage for ideals or line bundles to adition langage for divisors: if you write proof of what you say, no theorem or difficult argument is necessary. But thank you for answer any way.

http://mathoverflow.net/questions/44005/intuition-for-rational-functions/44040#44040 evgeniamerkulova 2010-10-29T07:24:54Z 2010-10-29T07:24:54Z

&quot;Computing the fiber of the field of all rational functions at a non-generic point likely has no classical counterpary&quot; is perfect answer to question and explain why I had not intuition for this concept. Thank you very much.

http://mathoverflow.net/questions/44005/intuition-for-rational-functions/44010#44010 evgeniamerkulova 2010-10-28T22:06:57Z 2010-10-28T22:06:57Z

Nobody uses the notation $\mathcal O(y)$ for regular functions vanishing at $y$: where have you seen that notation? If $X$ is a curve it is very confusing because one allways uses it for sheaf of rationnal functions having pole (not zero!) of order $\leq 1$ on $y$ and regular outside $y$.