4 added 57 characters in body

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 themthese 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 on 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 primesprime 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).

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 which involves at least $2$ parameters. Pick one of them, 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 on $(I,x)$ 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 parameters generate primes. 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).

2 added 1237 characters in body

ADDED: here is a proof of the statement you need 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 which involves at least $2$ parameters. Pick one of them, 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 on $(I,x)$ 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.

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