@Hailong: That's quite slick. Thanks a lot for taking all the time and effort.

@Hailong: I missed your last comment here. The point of the proof is expressing an ideal generated by squarefree monomials as an intersection of its minimal primes. Then it would follow it is radical. A priori we don't know that.

contd: A regular local ring is a UFD, ideals generated by subsets of a regular system of parameters are prime, so I thought the same proof would go through if we can push modularity in some form. I think it would be sufficient if modularity holds when $J$ is a parameter ideal.

@Hailong: Sorry, I was under the impression that the proof was standard in the polynomial case. Essentially the monomial ideal is expressed as an intersection of its minimal primes which are ideals generated by subsets of the indeterminates. The main ingredient of the proof are (1) that intersection distributes over sum when all ideals are monomial (modularity) and (2) the intersection of principal ideals is a principal ideal generated by the lcm of the generators in a UFD.

@Hailong: Thanks for your answer and the link. I eventually figured out modularity for parameter ideals. However, (as I posted in my follow up comment above), this is perhaps not sufficient for the proof to go through. Its certainly sufficient if modularity holds for ideals generated by monomials on a regular system of parameters.

In the above comment I mean, "if the modularity law holds for ideals generated by squarefree monomials on a regular system of parameters".

@Hailong: By modularity law, I mean for any if $J,K,L$ are ideals, then $J\cap (K+L)=J\cap K + J\cap L$. This is the terminology used in Atiyah-Macdonald (and some set theory books). If this holds for ideals generated by subsets of a regular system of parameters, then, I think, we are done. By "I am not sure I see where the problem is", do you mean you know/think this is true?

@Hailong: Thanks for the answer. Just to make it completely explicit. If $P$ is a complete intersection in a regular local ring (hence Cohen Macaulay), there is a regular sequence that generates it. We can reduce to case where $P$ is $2$ generator using induction. Then, using your claim and the fact that regular sequences are permutable in a Noetherian local ring, we get the result. Does this sound OK?

@Karl: Yes, definitely a more compact (better) way of putting it.