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Let $R = k[x, y]$ with $k$ algebraically closed, and $\mathfrak m = (x, y)$. Suppose $I$ is an $\mathfrak m$-primary ideal of $R$, i.e., $(x, y)^n \subset I \subset (x, y)$ for some $n$. If $I_{\mathfrak m}$ is generated by a regular sequence of length $2$, i.e., $I_{\mathfrak m} = aR_{\mathfrak m} + bR_{\mathfrak m}$ where $a, b$ is a regular sequence of $R_{\mathfrak m}$, what can we say about the number of generators of $I$ in this case?

All my examples show that $I$ is generated by a regular sequence of length $2$, yet not a proof is found.


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up vote 1 down vote accepted

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).

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