Number of generators of $m$-primary ideals in $k[x, y]$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:09:01Z http://mathoverflow.net/feeds/question/100915 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100915/number-of-generators-of-m-primary-ideals-in-kx-y Number of generators of $m$-primary ideals in $k[x, y]$ Leslie Wu 2012-06-29T07:24:51Z 2012-06-29T13:34:30Z <p>Let $R = k[x, y]$ with $k$ algebraically closed, and $m = (x, y)$. Suppose $I$ is an $m$-primary ideal of $R$, i.e., $(x, y)^n \subset I \subset (x, y)$ for some $n$. If $I_m$ is generated by a regular sequence of length 2, i.e., $I_m = aR_m + bR_m$ where $a, b$ is a regular sequence of $R_m$. What can we say about the number of generators of $I$ in this case?</p> <p>All my examples show that $I$ is generated by a regular sequence of length 2, yet not a proof is found. </p> <p>Thanks, </p> http://mathoverflow.net/questions/100915/number-of-generators-of-m-primary-ideals-in-kx-y/100931#100931 Answer by Mohan for Number of generators of $m$-primary ideals in $k[x, y]$ Mohan 2012-06-29T13:34:30Z 2012-06-29T13:34:30Z <p>This is a standard result using elementary homological algebra. If $I$ is a height two local complete intersection ideal in $R=k[x,y]$, then it is a complete intersection. Under the hypothesis, it follows that $\mathrm{Ext}^1_R(I,R)$ is isomorphic to $R/I$, this being a local calculation and Chinese remainder theorem. The extension corresponding to $1\in R/I$ is $0\to R\to P\to I\to 0$ and one checks that $P$ is $R$-projective of rank two, since by choice $\mathrm{Ext}^1_R(P,R)=0$, and hence free (by Seshadri's theorem). </p>