Are squarefree monomial ideals on a regular system of parameters in a regular local ring radical? Suppose $(R,m)$ is a regular, local ring. Let $x_1,x_2,...,x_n$ be a regular system of parameters. Let $I$ be an ideal generated by squarefree monomials in the $x_i$'s. Is $I$ a radical ideal? The motivation for this is the polynomial ring in finitely many indeterminates over a field (which although not local is regular) and the indeterminates generate the homogeneous maximal ideal. Then every ideal generated by squarefree monomials in the indeterminates is radical. 
I was trying to see whether the proof from monomial case in polynomial rings carries over. There we express an ideal generated by squarefree monomials as an intersection of ideals generated by subsets of the indeterminates. Each such ideal is prime. In the case of a regular system of parameters in a regular local ring, any subset of a rsop's also generates a prime ideal. So the only obstruction in the proof is whether the modularity law holds for such ideals. 
 A: ADDED: here is a proof of the statement you need (namely the square free monomial ideal $I$ is a intersection of primes generated by subsets of parameters) without using the modularity property. We will use induction on $N=$ the total numbers of times the parameters appear in the generators of $I$. For example if $I=(xy, xz)$ then $N=4$. The statement is obvious if $N=1$. 
Suppose $I$ has a generator (say $f_1$) which involves at least $2$ parameters. Pick one of these parameters, say $x$ and WLOG, we can assume $I=(f_1,\cdots, f_n, g_1,\cdots,g_l) $ such that $x|f_i$ for each $i$ but $x$ does not divide any of the $g_j$s. Let $F_i=f_i/x$. We claim that:
$$  I = (I,x) \cap (I,F_1)$$   
If the claim is true, we are done by applying the induction hypothesis to $(I,x)$ and $(I,F_1)$. One containment is obvious, for the other one we need to show if $xu \in (I,F_1)$ then $xu\in I$. 
Write $$xu = f_2x_2 + \cdots f_nx_n + \sum g_jy_j + F_1x_1$$ which implies
$$x(u- F_2x_2 +\cdots F_nx_n) \in (g_1,\cdots, g_l, F_1) = I' $$
$I'$ has minimal generators which do not contain $x$. By induction, $I'$ is an intersection of primes generated by other parameters, so $x$ is a NZD on $R/I'$. So $(u- F_2x_2 +\cdots F_nx_n) \in I'$, and therefore $xu \in I$, as desired. 
REMARK: note that for this proof to work, you only need that all subsets of the sequence (not necessarily parameters) generate prime ideals. I guess it fits with your other question. 

So from the comments I will take your question as proving $J\cap (K+L) = J\cap K + J \cap L$ for parameter ideals (by which I mean ideals generated by subsets of a fixed regular s.o.p). 
It will suffice to understand $I\cap J$ for two such ideals. To be precise, let $g(I)$ be the set of s.o.p generators of $I$. Let $P$ be the ideal generated by the intersection of $g(I),g(J)$, and $I', J'$ generated by $g(I)-g(P), g(J)-g(P)$. Then we need to show:
$$I \cap J  = P + I'J' $$
Since $R/P$ is still regular we can kill $P$ and assume that $g(I), g(J)$ are disjoint, and we have to prove $I \cap J = IJ$. This should be an easy exercise, but a slick and very general way is invoking Tor (which shows that  this is even true for $I,J$ generated by parts of a fixed regular sequence).
