User knot - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:56:55Z http://mathoverflow.net/feeds/user/27041 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125814/maximal-chain-of-1s-in-binary-strings Maximal chain of 1s in binary strings Knot 2013-03-28T11:10:42Z 2013-03-28T15:20:46Z <p>Let $S$ be the set of $2^n$ binary $n$-bit strings. For every $x\in S$, let $f(x)$ is the maximal chain of bits 1 in $x$. So Can we find a good upper bound of $$F(n)=\frac{\sum_{x\in S}f(x)}{2^n}$$ Of course, $O(1)\le F(n) \le O(n)$. I think the upper bound is a constant or $O(\log n)$. Can anyone help me?</p> http://mathoverflow.net/questions/108973/homogeneous-ideal-and-its-system-of-generators Homogeneous ideal and its system of generators Knot 2012-10-06T02:47:50Z 2013-03-13T20:45:03Z <p>Let $I$ be a homogeneous ideal in a graded commutative ring $R$, $S$ be its minimal system of generators.</p> <p>What is the conclusion that we can say about the element in $S$ ? Is the cardinality of $S$ uniquely determined by $I$ ? </p> <p>In the book Commutative ring theory of Matsumura, theorem 2.3, page 8 there is a theorem for local ring which we can deduce that the number of generator is unique. So, is there a version of that theorem for the graded ring ?</p> <p>What about the degree of generator in $S$ ? </p> http://mathoverflow.net/questions/121951/combinatorial-inequality Combinatorial Inequality Knot 2013-02-15T22:12:47Z 2013-02-18T15:48:38Z <p>Consider a set of $2^n-1$ non negative integers $S=${$a_{i,j}|1\le i\le n; 1\le j\le 2^{i-1}$} such that:</p> <p>\begin{align}{} 1.\ \ &amp;a_{i,j}\le 2^{n+1-i} \\ 2.\ \ &amp;a_{i,j}\le a_{i-1,j} \\ 3.\ \ &amp;a_{1,1}+\sum_{i=2}^n\left(\sum_{j=2^{i-2}+1}^{2^{i-1}}{a_{i,j}}\right)\le 2^n \end{align} Prove that: $$\sum_{S}a_{i,j}\le k2^n$$ in which $k$ is a constant number.</p> http://mathoverflow.net/questions/111392/on-the-paper-on-the-asymptotic-linearity-of-castelnuovo-mumford-regularity On the paper "On the asymptotic linearity of Castelnuovo-Mumford regularity" Knot 2012-11-03T16:44:44Z 2012-11-03T19:32:18Z <p>I have posted this question on MSE, however it seems not to be interested by member there, so I decided to post it here. I am sorry if you feel it is not appropriate for MO.</p> <p>I am now reading the paper on Castelnuovo-Mumford regularity of Ngo Viet Trung and Hsin-Ju Wang, which can be found <a href="http://arxiv.org/pdf/math/0212161.pdf" rel="nofollow">here</a></p> <p>In this paper, the authors gave the concept of flat extension of ring, and in my opinion, they gave an example of it (<em>in the lemma 1.2</em>): </p> <blockquote> <p>Let $A$ be a Noetherian ring, $R$ be standard graded finitely generated $A$-algebra, $R_{+}$ be its irrelevant ideal, $M$ be a finitely generated $R$-module. Let $Q$ be a $M$-reduction of $R_+$, generated by linear form $x_{1},...,x_{s}$. For $i=1,...,s$, put $z_{i}=\sum_{j=1}^{s}u_{ij}x_{j}$ where $U=\lbrace u_{ij}|i,j=1,...,s\rbrace$ is a matrix of indeterminates. Put $A'=A[U,\text{det}(U)^{-1}], R'=R\otimes_{A}A', M'=M\otimes_{A}A'$ then $A'$ is a flat extension of $A$.</p> </blockquote> <p>As I understand $A'$ is a flat extension of $A$ if regarded as $A$ module, tensoring a exact sequence with $A'$ will give us an exact sequence.My questions are</p> <ol> <li>How can we prove that the ring $A'$ constructed above satisfies the above property ? </li> <li>From a commutative ring $A$, how can we construct its flat extension ?</li> </ol> <p>I also have questions in the proof of the proposition 2.1 in that paper</p> <blockquote> <p>Let $S=A[X_{1},...,X_{S}, Y_{1},...,Y_{v}]$ be a polynomial ring over a commutative Noetherian ring $A$. This ring can be viewed as a bigraded ring if we associate the degree for $X_{i}$ and $Y_{j}$ as $\text{deg}(X_{i})=(1,0), i=1,...,s; \text{deg}(Y_{j})=(d_{j}, 1), j=1,...,v$</p> <p>Now, let $\mathcal{M}$ be a finitely generated graded module over <code>$A[X_{1},...,X_{s}, Y_{1},]$</code>. For a fixed number $n$, put <code>$\mathcal{M}_{n}=\oplus_{a\ge0}\mathcal{M}_{(a,n)}$</code>. Then <code>$\mathcal{M_n}$</code> is a finitely generated graded module over the naturally graded polynomial ring $A[X_1,...,X_s]$. If $s=0$ then <code>$\text{reg}(\mathcal(M_{n})$</code> is the invariant : $a(\mathcal{M_{n}})=\text{max}\lbrace a|\mathcal{M_{(a,n)}}\neq 0\rbrace$</p> </blockquote> <p>3-Question : Why $\text{reg}\mathcal{M_n})$ is equal to $a(\mathcal{M_n})$ ? </p> <p>As I understand , in $\mathcal{M_{(a,n)}}$, the index $a$ is graded for the variables $X_i$ and the index $n$ is graded for the variables $Y_{J}$ and then $\mathcal{M_{n}}$ is graded module over $A[X_1,…,X_s]$ but it is not a graded module over $A[Y_1,…,Y_v]$. So if $s=0$, and $n$ is fixed then we get nothing on $\mathcal{M_n}$. So, what is the point here ?</p> <p>Proposition 2.1 in that paper says that </p> <blockquote> <p>Let $\mathcal{M_n}$ be a finitely generated bigraded module over the bigraded ring $A[Y_1,...,Y_v]$. Then $a(\mathcal{M_n})$ is asymptotically a linear function with slope $\leq d_v$</p> </blockquote> <p>Here is the argument in the proof :</p> <blockquote> <p>The case $v=0$ is trivial.Consider the exact sequence : <code>$0\rightarrow [0_{\mathcal{M}}:Y_{v}]_{(a,n)}\rightarrow \mathcal{M}_{(a,n)}\xrightarrow{Y_{v}}\mathcal{M}_{(a+d_{v},n+1)}\rightarrow \mathcal{M/Y_{v}M}_{(a+d_{v},n+1)}\rightarrow 0$</code></p> <p>Since $0_{\mathcal{M}}:Y_{v}$ and $\mathcal{M}/Y_{v}\mathcal{M}$ can be viewed as bigraded modules over $A[Y_1,...,Y_{v-1}]$, using induction we may assume that <code>$a([0_{\mathcal{M}}:Y_{v}]_{n})$</code> and <code>$a([\mathcal{M}/Y_{v}\mathcal{M}]_{n})$</code> are asymptotically linear functions with slope $\leq d_v$. As consequence : <code>$$a([0_{\mathcal{M}}:Y_{v}]_{n})+d_v\ge a([0_{\mathcal{M}}:Y_{v}]_{n+1})$$</code> <code>$$a([\mathcal{M}/Y_{v}\mathcal{M}]_{n})+d_v\ge a([\mathcal{M}/Y_{v}\mathcal{M}]_{n+1})$$</code> for large $n$.</p> </blockquote> <p>My question for this part is :</p> <p>4- How did the author use the induction ? If they use the induction on $v$ then why can they get the above two inequalities, since I think we can get the inequalities if we use induction on $n$.</p> <blockquote> <p>Since <code>$a(\mathcal{M}_{n})\ge a([0_{\mathcal{M}}:Y_{v}]_{n})$</code> for all $n$, we only need to consider the case that <code>$a(\mathcal{M}_{m})&gt; a([0_{\mathcal{M}}:Y_{v}]_{m})$</code> for infinitely many $m$. Putting this condition into the above exact sequence we get : <code>$$a(\mathcal{M}_{m+1})=\text{max}\lbrace a(\mathcal{M}_{m})+d_{v},a([\mathcal{M}/Y_{v}\mathcal{M}]_{n+1})\rbrace$$</code></p> </blockquote> <p>My question for this part is : </p> <p>5-How can the author get the above information from the exact sequence ?</p> <p>Here is my argument : We have : <code>$Y_{v}\mathcal{M}_{n}=\oplus_{a\ge 0}Y_{v}\mathcal{M}_{(a,n)}:=\oplus_{a\ge 0}\mathcal{N}_{(a+d_{v},n+1)}$</code>. Then : <code>$$a(Y_{v}\mathcal{M}_{n})=a(\oplus_{a\ge 0}\mathcal{N}_{(a+d_{v},n+1)})$$</code> <code>$$=\text{max}\lbrace a+d_{v}|\mathcal{N}_{(a+d_{v},n+1)}\neq 0\rbrace$$</code> <code>$$=\text{max}\lbrace a+d_{v}|Y_{v}\mathcal{M}_{(a,n)}\neq 0\rbrace$$</code> <code>$$=d_{v}+\text{max}\lbrace a|\mathcal{M}_{(a,n)}\neq 0\rbrace$$</code> <code>$$=d_{v}+a(\mathcal{M_{n}})$$</code> Since <code>$Y_{v}\mathcal{M}_{n}\subseteq\mathcal{M}_{n+1}$ then $a(Y_{v}\mathcal{M}_{n})\le a(\mathcal{M}_{n+1})$</code> or <code>$a(\mathcal{M}_{n+1})\ge d_{v}+a(\mathcal{M}_{n})$</code>. </p> <p>6-Am I wrong anywhere ? </p> <p>7-What is the aim of the authors in the rest of the proof ?</p> <p>I know that my questions are very long and may be hard to follow. If you have any question on this or this kind of questions are not appropriate with MO please let me know.</p> <p>Thank for reading!</p> http://mathoverflow.net/questions/110381/filter-regular-sequence-and-regularity Filter-regular sequence and regularity Knot 2012-10-23T02:05:56Z 2012-10-23T13:32:21Z <p>Let $A$ be a commutative Noetherian ring, $R$ be a standard graded algebra over $A$, $M$ be finitely generated graded $R$-module. Let $R_{+}$ be the irrelevant ideal. The Castelnuovo-Mumford regularity of $M$ or regularity for short is defined to be : $\text{reg}(M):=\text{max}\lbrace a(H_{R_+}^{i}(M))+i|i\ge 0\rbrace$. Let $x_1,...,x_s$ be linear form in $R$. This set of elements is called a $M$-filter regular sequence if $x_{i}\notin \mathfrak{p}$ for any associated prime $\mathfrak{p}\nsupseteq R_{+}$ of $(x_1,...,x_{i-1})M$ for $i=1,...,s$</p> <p>Then, people claim that : $\text{reg}(M)=\text{max}\lbrace a((x_1,...,x_i)M:R_{+}/(x_1,...,x_{i})M)|i=1,...,s\rbrace$</p> <p>Could you explain for me why is it ? What is the motivation of filter regular sequence ?</p> http://mathoverflow.net/questions/109672/question-on-bigraded-modules Question on bigraded modules Knot 2012-10-15T03:16:27Z 2012-10-15T22:00:45Z <p>Let $R$ be a polynomial ring $k[x_1,...,x_n]$, let $f_1,...f_s$ be some non-zero polynomial sin $R$ of degree $p_1,...,p_s$ respectively.Define $S$ by $k[X_1,...,X_n, T_1,...T_s]$ with bigrading defined by $degX_i=(1,0)$ and $degT_j=(p_j,1)$. For a bigraded $S$-module $M=\bigoplus_{p,n\in\Bbb{N}}M_{(p,n)}$, define $M^{(n)}$ to be the graded $R$-module $\bigoplus_{p\in\Bbb{N}}M_{(p,n)}$ in which the action of $x_i$ can be understanded as $X_i$ with its obvious grading.</p> <p>There are two claims that I could not prove, so I decided to post it here, and I hope you will help me to prove it.</p> <p>1-The functor $(.)^{(n)}$ is an exact functor.</p> <p>2-There are the isomorphisms: $S(-a,-b)^{(n)}\cong S^{(n-b)}(-a)\cong \bigoplus_{a_1+...+a_s=n-b}R{(-a_1p_1-...-a_sp_s-a)}$</p> <p>Thank for reading my question!</p> http://mathoverflow.net/questions/109194/on-the-generator-of-power-of-ideal On the generator of power of ideal Knot 2012-10-09T01:07:29Z 2012-10-13T15:20:09Z <p>Let $I$ be a graded ideal in a polynomial ring $R$, which is generated minimally by $x_1,...,x_k$. Then the power of $I$, i.e $I^t$ is generated by monomials of the form $x_{1}^{a_1}...x_{n}^{a_{n}}$ where $a_1+...+a_n=t$. Denote this set by $S$.</p> <p>Can we say anything (others than above)about the minimal generating set of $I^{t}$? Is it $S$ ?</p> <p>Given a minimal generating set for a graded ideal in a graded commutative ring, from these how much do we know about the minimal generating set for the power of it?</p> <p><strong>Edit</strong> : Here is an example for precising my question :</p> <p>In the polynomial ring $k[x,y,z]$ let $I=(x^2, xy^3, y^2z^3)$, then $I^2=(x^4, x^2y^6, y^4z^6, x^3y^3, xy^5z^3, x^2y^2z^3)$</p> <p>Is $\lbrace x^4, x^2y^6, y^4z^6, x^3y^3, xy^5z^3, x^2y^2z^3\rbrace$ a minimal generating set for $I^2$ ?</p> <p><strong>Update</strong> There are some typing mistake that I have not noticed. I have change my question. This time, the generating set of $I$ is minimal. So what can we say about the generating set for $I^2$ above ? Is it minimal? Thank you everyone for helping me answer my question!</p> http://mathoverflow.net/questions/109139/castelnuovo-mumford-regularity-and-degree-of-generator Castelnuovo-Mumford regularity and degree of generator. Knot 2012-10-08T12:30:46Z 2012-10-09T04:57:23Z <p>Let $R$ be a polynomial ring over a field $k$,: $k[x_{1},..x_{n}]$, $\mathfrak{m}=(x_0,...,x_{n})$ and $M$ be a finitely generated $R$ module.</p> <p>In a paper of Kodiyahlam, he define the Castelnuovo-Mumford regularity of $M$ to be the least integer number $m$ such that for every $j$ the $j$-th syzygy of $M$ is generated in degree less or equal $m+j$.Then, he conclude that the Castelnuovo-Mumford regularity of $M/\mathfrak{m}M$ is equal to the maximal degree of generator of $M$.</p> <p>Could you please so me why the CM regularity of $M/\mathfrak{m}M$ is equal to the maximal degree of generator of $M$ ? </p> <p>If $I$ and $J$ are two finitely generated ideal of $R$ and $J\subseteq I$ then do we have the CM regularity of $J$ is less or equal the CM regularity of $I$ ?</p> http://mathoverflow.net/questions/125814/maximal-chain-of-1s-in-binary-strings Comment by Knot Knot 2013-03-28T13:21:14Z 2013-03-28T13:21:14Z @ Noah Stein. &quot;the&quot; upper bound is O(1), O(log n), O(\sqrt(n)) or O(n)... We do not need the exact value of F(n). http://mathoverflow.net/questions/125814/maximal-chain-of-1s-in-binary-strings Comment by Knot Knot 2013-03-28T13:17:47Z 2013-03-28T13:17:47Z And thank for your useful suggestion, Nadeau. :) http://mathoverflow.net/questions/125814/maximal-chain-of-1s-in-binary-strings Comment by Knot Knot 2013-03-28T13:05:06Z 2013-03-28T13:05:06Z I'm sorry, this is a typo. Yeah, it is 2^n at the denominator http://mathoverflow.net/questions/121951/combinatorial-inequality Comment by Knot Knot 2013-02-17T09:39:25Z 2013-02-17T09:39:25Z It seems to me that $\frac{\sum a_{ij}}{2^n}=O(\log n)$. Is it correct? http://mathoverflow.net/questions/111392/on-the-paper-on-the-asymptotic-linearity-of-castelnuovo-mumford-regularity Comment by Knot Knot 2012-11-05T04:17:46Z 2012-11-05T04:17:46Z @Graham Leuschke : Thanks ! http://mathoverflow.net/questions/111392/on-the-paper-on-the-asymptotic-linearity-of-castelnuovo-mumford-regularity/111403#111403 Comment by Knot Knot 2012-11-05T04:17:29Z 2012-11-05T04:17:29Z @William Sawin : Thanks. I really appreciate that. What about the other questions ? http://mathoverflow.net/questions/109672/question-on-bigraded-modules/109766#109766 Comment by Knot Knot 2012-10-16T01:50:07Z 2012-10-16T01:50:07Z @Ralph: Thank you very much for answering my question. May I ask : Why in $(*)$ the component containging the variables $T1,...T_s$ disappear? http://mathoverflow.net/questions/109672/question-on-bigraded-modules Comment by Knot Knot 2012-10-15T13:58:10Z 2012-10-15T13:58:10Z @Ralph: You are right. In fact, I proved it but I am not sure about it, the graded structure confuses me. http://mathoverflow.net/questions/109194/on-the-generator-of-power-of-ideal/109431#109431 Comment by Knot Knot 2012-10-15T13:43:41Z 2012-10-15T13:43:41Z @Neil Epstein : What will happen if I is a monomial ideal ? http://mathoverflow.net/questions/109672/question-on-bigraded-modules Comment by Knot Knot 2012-10-15T13:40:56Z 2012-10-15T13:40:56Z @Ralph: Thank you, I have confused with $R$ and $S$. The 2nd question is in fact a results of a lemma in a paper, that I do not think it is trivial but I have not proved it yet, so I decided to post it here. Thank you very much! P.s : The functor is from the category of bigraded modules to the category of bigraded module, I think. http://mathoverflow.net/questions/109672/question-on-bigraded-modules Comment by Knot Knot 2012-10-15T03:56:24Z 2012-10-15T03:56:24Z @Mariano Suarez-Alvarez : Thank you very much! I edited it. http://mathoverflow.net/questions/109194/on-the-generator-of-power-of-ideal/109431#109431 Comment by Knot Knot 2012-10-13T15:24:14Z 2012-10-13T15:24:14Z Dear Hailong Dao, firstly I want to say thank for your answer, it helps me alot. Secondly, I do not care about the fact that the question may let people think I am stupid, I just want to know the answer and how to think about it, it is more worthy for me. About your answer, can I ask(may be another stupid one) a question : If $x_1,..x_n$ minimally generate a graded ideal $I$, and they form a regular sequence, then S is the minimal generating set ? http://mathoverflow.net/questions/109194/on-the-generator-of-power-of-ideal Comment by Knot Knot 2012-10-11T23:34:31Z 2012-10-11T23:34:31Z Sorry everybody for my stupid mistake in the example. I have edited it. Please help me. Thanks. http://mathoverflow.net/questions/109383/algorithm-for-computing-the-reduction-of-a-graded-ideal Comment by Knot Knot 2012-10-11T17:18:20Z 2012-10-11T17:18:20Z @Mahdi: I just want to apply the algorithm in a very special case : Computing the reduction that I need right now, and I do not have much time to think about the other things :(. Btw, thank you very much! http://mathoverflow.net/questions/109383/algorithm-for-computing-the-reduction-of-a-graded-ideal Comment by Knot Knot 2012-10-11T16:27:29Z 2012-10-11T16:27:29Z @Dima Pasechnik : Could you please tell me which book that I can read for applying Grobner basis to compute the reduction ?