a "paradox" regarding regular and complete intersection rings 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? 
 A: 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.
A: 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.
