# Example of height $n$ ideal with $I/I^2$ (locally) $n$-generated, but $I$ is not.

For $R$, a commutative noetherian ring of dimension $d$, I'm looking for an example where $I \subset R$ is an ideal of height $n \lt d$ such that $I/I^2$ is generated by $n$ elements (locally $n$-generated is also fine), however, $I$ itself is not. Moreover, it would be greatly helpful if your response could address the geometric intuition of the example as well.

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Have you considered a non-regular local ring $R$ with maximal ideal $I$? Then I suspect $I\subset R[x]$ would suffice (as this fulfills the additional requirement $n<d$), but someone should check me on this. –  Alex Becker Feb 19 '12 at 23:35

Alex is right. Let $R=(\mathbb Z[\sqrt{6}])[X]$, and take $I=(2,\sqrt{6})$. Then $R$ has dimension $2$, $I$ has height $1$ but is generated by two elements, and $I^2=(2,2\sqrt{6})$ so $I/I^2$ is generated by the element $\sqrt{6}+I^2$.

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Take a Dedekind domain with a non principal prime ideal $P$. Then $P/P^2$ is generated by one element but not $P$.

Your question has negative answer because you ask too much. However, under your hypothesis $I$ is locally (on $\mathrm{Spec}(R)$) generated by $n$ elements (use Nakayama).

EDIT I said "you ask too much" because the geometric interpretation of your hypothesis on $I/I^2$ is that there exists a homomorphism $O_X^n \to I^{\sim}$ of $O_X$-modules (where $X=\mathrm{Spec}(R)$, $O_X$ is the structural sheaf and $I^{\sim}$ is the coherent sheaf associated to $I$) which is surjective at points of $V(I)$ (so surjective in some open neighborhood of $V(I)$). But there is no reason that the surjectivity extends to the whole $X$.

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I'm not clear on your second sentence... It is totally fine with me if $I$ is locally generated by the same number of elements as $I/I^2$, so long as it is not globally so - in fact, that's even better. –  Andrew Parker Feb 23 '12 at 5:24
@AndrewParker: take a point $x\in X:=\mathrm{Spec}(R)$. Then the stalk $I_x/I_x^2$ is generated by $n$ elements ($I_x$ is the ideal of $O_{X,x}$ generated by $I$). If $x\notin V(I)$, then $I_x=O_{X,x}$ and $I$ is generated by $1$ in a neighborhood of $x$. Suppose $x\in V(I)$. Then $I_x\subseteq m_x$ the maximal ideal of $O_{X,x}$. So $I_x/m_xI_x$, being a quotient of $I_x/I_x^2$, is also generated by $n$ elements. By Nakayama, $I_x$ is generated by $n$ elements. If you need further details, I will put them in the answer. –  user18119 Feb 24 '12 at 12:49
Yes, please. The geometric perspective is exactly what I'm after. Though, from your comment, it seems as though you're still saying that $I$ is locally $n$-generated (which is fine), and I know that this does not necessarily mean that $I$ is $n$-generated. What I wasn't clear on was "your question has a negative answer because you ask too much". I was looking for an example where $I$ was locally n-generated but not globally so. –  Andrew Parker Feb 24 '12 at 14:44