On the generator of power of ideal - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:00:52Z http://mathoverflow.net/feeds/question/109194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf 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/109194/on-the-generator-of-power-of-ideal/109431#109431 Answer by Hailong Dao for On the generator of power of ideal Hailong Dao 2012-10-12T05:31:03Z 2012-10-12T05:31:03Z <p>I have to admire your persistence, perhaps you really want an answer (-: </p> <p>In general, the answer to your first question (second paragraph) is NO, it is not $S$, even for monomial ideals in a polynomial rings. Take the ideal $I$ generated by $x_1 = a^4b, x_2=b^4a, x_3=a^3b^3$. Then $x_3^2$ is not a minimal generator for $I^2$ since it is divisible by $x_1x_2$.</p> <p>One particular case when the answer is YES is when the $x_1,\cdots, x_n$ form a <em>regular sequence</em>. </p> <p>For specific examples, it may be worth learning some program such as Macaulay 2. In your specific example $S$ is the minimal generating set for $I^2$, provided that you fix the fifth entry as Gerhard pointed out. </p> <p>For a monomials ideal one can also visualize the minimal generators as the points on the convex hull of the set of degrees of the ideal. </p> <p>As my first example shows, one can not make very good statement about a specific power of $I$. However, <em>asymptotically</em> we can say quite a bit:</p> <blockquote> <p>The minimal number of generators of $I^n$ for $n>>0$ is a polynomial in $n$. The degree of this polynomial is called the analytic spread of $I$ (geometrically it is one more than the dimension of the exceptional fibre of the blow-up). </p> </blockquote>