The following "paradox" arose as i was studying the proof of Theorem 2.3.3 in Bruns and Herzog, CMR. My question is self-contained but i could expand on details upon request.

Let $(S,\mathfrak{n})$ be a regular local ring and let $I$ be an ideal of $S$ such that $I \subset \mathfrak{n}^2$. Now suppose that $R=S/I$ is a complete intersection ring. Then $\mu(I) = \operatorname{emb.dim} R - \dim R$, where $\mu(I)$ is the minimal number of generators of $I$ (e.g. Matsumura, CRT, p. 170-171 or Bruns and Herzog Theorems 2.3.2(b) and 2.3.3(b)). Because $I \subset n^2$, we have that $\operatorname{emb.dim} R = \dim S$. Hence $\mu(I) = \dim S - \dim S/I$. But $S$ is local regular and hence Cohen-Macaulay, and so $\operatorname{height} I + \dim S/ I = \dim S$, thus $\mu(I) = \operatorname{height}(I) =: \nu$. So there exist elements $x_1, \cdots, x_{\nu}$ inside $I$ such that $I = (x_1,\dots,x_{\nu})$. It follows that $\dim S/(x_1,\dots,x_{\nu}) = \dim S - \nu$ and since $S$ is regular, we conclude that $(x_1,\dots,x_{\nu})$ is a part of a system of parameters or equivalently that the images of $x_1,\dots,x_{\nu}$ mod $\mathfrak{n}^2$ are linearly independent. But by hypothesis $I \subset \mathfrak{n}^2$ and so every $x_i$ maps to zero in $\mathfrak{n}/ \mathfrak{n}^2$, CONTRADICTION.

Question: Where is the mistake in the above argument

********* EDIT ******************

After Daniele's comment, let me present essentially the above argument more directly:

Let $R$ be a complete intersection ring. Hence there exists a regular local ring $(S,n)$ and an $S$-sequence $x :=x_1,\dots,x_{\nu}$ such that $R=S/(x)$. Since $S$ is regular, then it is Cohen-Macaulay and so by Theorem 2.1.2 in Bruns and Herzog, CMR, $x$ is part of a system of parameters of $S$. Then by Theorem 14.2 in Matsumura, CRT, $S/(x)=R$ will be regular.

Now, the only thing that i assumed above is that for a regular local ring $S$ "a system of parameters" is the same thing as "a regular system of parameters" and i now see that this may not be the case. Is that it?

  • $\begingroup$ I think that the problem is that $S/(x_1,\dots,x_\nu)$ will not be regular, so that it is not true that the $x_1,\dots,x_\nu$ can be extended to a system of parameters. For example, if $S=k[X]_{(X)}$ and $I=(X^2)$ then $S/(X^2)$ is not regular since it is not reduced. I think that your argoment proves the general fact by contradiction. $\endgroup$ – Daniele A Feb 7 '14 at 19:45
  • $\begingroup$ @DanieleA: I am sure see your point: Since $S$ is regular and since i proved that $x_1,\dots,x_{\nu}$ is an $S$-sequence (equivalently a part of system of parameters), then $S/(x_1,\dots,x_{\nu})$ has to be regular (which i think is another contradiction, maybe that's what you are saying). But anyway, i am looking to find where is the flaw in the above argument. $\endgroup$ – Manos Feb 7 '14 at 19:55
  • $\begingroup$ @DanieleA: Even though i am still not entirely convinced, i think you are almost right: $x_1,\dots,x_{\nu}$ is part of a system of parameters, however it may not be necessary that it is a part of a REGULAR system of parameters. $\endgroup$ – Manos Feb 7 '14 at 20:41
  • $\begingroup$ yes, that's precisely the problem with the argument, as it has been clearly explained by user115654 below. $\endgroup$ – Daniele A Feb 9 '14 at 14:54

As has been pointed out, the difference is between a system of parameters and a regular system of parameters. An sop is just a set of $\dim R$-many elements which generate an $m$-primary ideal, whereas a regular sop is also required to be a set of minimal generators of $m$, or equivalently by Nakayama, they form a basis of $m/m^2$. Regular sop's are sop's, but they only exist if the ring is regular.

In the case of your question, $R$ is a complete intersection, so $R$ has a presentation of the form $S/(x)$ where $S$ is regular local and $(x)$ is an $S$-regular sequence (if $R$ is complete). It is true that in a Cohen-Macaulay local ring, a regular sequence can be extended to an sop, but even though $S$ is regular, it need not extend to a regular sop - this is the case precisely when $R$ is regular.

  • $\begingroup$ Thank you for your very clear answer :) $\endgroup$ – Manos Feb 8 '14 at 2:03
  • $\begingroup$ @Manos: You're very welcome $\endgroup$ – zcn Feb 8 '14 at 2:11

There is an error at the end when you say: "$\ldots$ or equivalently that the images of $x_1\ldots,x_\nu$ mod $\mathfrak{n}^2$ are linearly independent." I don't see why this is equivalent to $(x_1,\ldots,x_\nu)$ being a system of parameters. For example suppose $y_1,\dots,y_\nu$ are a set of generators of the maximal ideal $\mathfrak{n}$. Then clearly $y_1^2,\dots,y_\nu^2$ generate a parameter ideal but their image mod $\mathfrak{n}^2$ is zero.

  • $\begingroup$ That sounds right. I was confused by the theorem that says that in a regular local ring $x_1,\dots,x_{\nu}$ is a part of a regular system of parameters if and only if their images mod $\mathfrak{n}^2$ are linearly independent. Any comments on that? $\endgroup$ – Manos Feb 7 '14 at 23:54
  • $\begingroup$ That is true, but as you wrote in one of your comments, not every system of parameters is a regular system of parameters. $\endgroup$ – Must Feb 8 '14 at 3:11

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