User nico bellic - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T22:24:03Z http://mathoverflow.net/feeds/user/12259 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116940/does-existence-of-an-isolated-solution-imply-the-jacobian-determinant-is-non-zero Does existence of an isolated solution imply the Jacobian determinant is non-zero? Nico Bellic 2012-12-21T01:34:52Z 2012-12-21T11:36:10Z <p>Let $f_1,\dots,f_n$ be formal power series in $\mathbb{C}[[x_1,\dots,x_n]]$ whose constant terms are all zero (i.e. $f_1,\dots f_n$ are not units in the ring). Suppose further that the radical of the ideal $(f_1,\dots f_n)$ is the maximal ideal. Does this imply that the Jacobian determinant $$\det \left(\frac{\partial f_i}{\partial x_j}\right)$$ is not identically zero?</p> <p><b>Note</b>, that in the special case when $f_1,\dots f_n$ are polynomials, we get the following statement: If the Jacobian determinant associated to a system of $n$ polynomial equations in $n$ variables with complex coefficients is identically zero, then the system has no isolated solutions.</p> http://mathoverflow.net/questions/92604/lengths-over-a-local-ring Lengths over a local ring Nico Bellic 2012-03-29T19:56:17Z 2012-04-17T19:07:11Z <p>Let $A$ be a noetherian domain, $\mathfrak{m}$ а maximal ideal, $s$ a non-zero element of $\mathfrak{m}$, $d= \dim A_\mathfrak{m}$. Is the following claim true?</p> <p>Claim: For any $\epsilon>0$, there exists a positive integer $n$ s.t. for any ideal $I$ satisfying</p> <p>1) $I \subset\mathfrak{m^n}$</p> <p>2) $\sqrt I = \mathfrak{m}$</p> <p>3) $I$ can be generated by $d$ elements,</p> <p>the following holds: $$\mbox{length}(A/(I+As)) /\mbox{length}(A/I) &lt; \epsilon$$</p> <p>Note: The following example shows that the claim can be false if one drops the requirement that that the number of generators of $I$ be bounded. </p> <p>Example: $A:= k[x,s]$, and let $\mathfrak{m}$ denote the ideal $(x,s)$. Let $I_{n,m}$ be an ideal of $A$ given by $$I_{n,m}= s\mathfrak{m}^{n-1} + \mathfrak{m}^m$$.</p> <p>We can calculate that for any $n$, $$\lim_{m\to \infty} \mbox{length}(A/(I_{n,m}+As)) /\mbox{length}(A/I_{n,m}) = 1$$ </p> http://mathoverflow.net/questions/55931/invariance-of-zx-under-a-self-equivalence-of-the-category-of-commutative-ring Invariance of $Z[x]$ under a self-equivalence of the category of commutative rings with 1. Nico Bellic 2011-02-19T00:02:37Z 2011-05-20T08:32:17Z <p>Let $\mbox{Rings}$ be the category of commutative rings with $1$. </p> <p>Is there an equivalence of categories $F: \mbox{Rings} \to \mbox{Rings}$ s.t.<br> $$F(\mathbb{Z}[x])\not\cong \mathbb{Z}[x]?$$</p> http://mathoverflow.net/questions/116940/does-existence-of-an-isolated-solution-imply-the-jacobian-determinant-is-non-zero Comment by Nico Bellic Nico Bellic 2012-12-22T20:01:57Z 2012-12-22T20:01:57Z @kreck: Can you offer any references that would enable me to understand your proof? http://mathoverflow.net/questions/92604/lengths-over-a-local-ring/94313#94313 Comment by Nico Bellic Nico Bellic 2012-04-20T03:46:00Z 2012-04-20T03:46:00Z Thanks a lot! I accept the answer. http://mathoverflow.net/questions/92604/lengths-over-a-local-ring Comment by Nico Bellic Nico Bellic 2012-03-30T22:08:32Z 2012-03-30T22:08:32Z Thank you for your comment! Maybe &quot; $I$ is parameter ideal&quot; can be replaced by &quot;number of generators of $I$ is bounded&quot;. I added an example that shows that some restriction on $I$ is necessary. http://mathoverflow.net/questions/20138/why-is-proj-of-any-graded-ring-isomorphic-to-proj-of-a-graded-ring-generated-in-d/20145#20145 Comment by Nico Bellic Nico Bellic 2012-01-12T04:03:14Z 2012-01-12T04:03:14Z Now we know how to solve that combinatorics problem ! :) http://mathoverflow.net/questions/55931/invariance-of-zx-under-a-self-equivalence-of-the-category-of-commutative-ring/56183#56183 Comment by Nico Bellic Nico Bellic 2011-02-22T22:11:24Z 2011-02-22T22:11:24Z Hi Martin, This is probably a stupid question, but looking at Lang didn't yield an answer. How do we know that there is no other purely transcendental, non-isomorphic to $\mathbb{Q}(x)$, extension of $\mathbb{Q}$ that also embeds inside every other transcendental extension of $\mathbb{Q}$? http://mathoverflow.net/questions/55931/invariance-of-zx-under-a-self-equivalence-of-the-category-of-commutative-ring/56183#56183 Comment by Nico Bellic Nico Bellic 2011-02-21T23:35:20Z 2011-02-21T23:35:20Z Hi Martin, Can you please explain more why $|R| \cong |F(R)|$ is a ring homomorphism? Also does it depend on the choice of an isomoprhism $\mathbb{Z}[x]\to F(\mathbb{Z}[x])$?