Are complete intersection prime ideals of regular rings regular ideals?

Let $(R, \mathfrak{m})$ be a regular local ring and let $\mathfrak{p}$ be a prime ideal of $R$ which is a complete intersection, i.e. the minimal number of generators of $\mathfrak{p}$ equals its height $h$. Then by Macaulays theorem there is a system of parameters (or equivalently - a regular sequence) $\{a_{1},\dots, a_{h}\}$ which generates $\mathfrak{p}$.

Is it then also true that $\mathfrak{p}$ can be generated by elements $\{b_{1}, \dots, b_{h}\}$ which can be extended to a regular system of parameters for $R$? Phrased differently, I am asking whether every complete intersection prime ideal in $R$ is regular (in the sense that $R/ \mathfrak{p}$ is regular).

I am asking this question being interested in the situation where $R = \mathbb{C}\{x_{1},\dots, x_{n}\}$ is the ring of convergent power series.

• Geometrically, you seem to be asking whether a complete intersection in a nonsingular affine variety is always nonsingular. It's not hard to guess that fails already in simple cases. mbrown's answer below (the cuspidal cubic) gives a concrete example.
– user64687
Oct 7 '13 at 13:56

Take $R=\mathbb{C}[x,y]_{(x,y)}$, and take $\mathfrak{p}=(x^2-y^3)$. $R/\mathfrak{p}$ is not a regular local ring, since it isn't integrally closed in its field of fractions.
• Thanks for the easy counterexample. Another way to see that $R/ \mathfrak{p}$ is not a regular local ring is by using completion and then the Jacobian criterium. The completion of $R/ \mathfrak{p}$ wrt $(x,y)$ is just $\hat{R}/ \hat{p} = \mathbb{C}[[x,y]]/ (x^{2}-y^{3})\cdot \mathbb{C}[[x,y]]$. By the Jacobian Criterium $\mathbb{C}[[x,y]]/ (x^{2}-y^{3})\cdot \mathbb{C}[[x,y]]$ is regular iff the Jacobian matrix of $x^{2}-y^{3}$ does not vanish at $(0,0)$. Oct 7 '13 at 15:48