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).
They are all equivalent, including definition 1, to the Krull dimension of $S/I$, where $S=\mathbb{R}[x_1,\ldots,x_n]$ is the polynomial ring $I$ lives in. This is very good news for people like me who want to apply algebraic geometry to statistics, where numbers are mostly real.
Here's how it goes:
Definition 1 is a way of computing the Krull dimension of $S/I$ via Groebner bases. See, e.g., p. 250 of Computing in algebraic geometry: a quick start using SINGULAR by Wolfram Decker and Christoph Lossen
Definition 2 is shown equivalent to Krull dimension in Corollary 2.8.9 of Real Algebraic Geometry by Bochnak, Coste and Roy. Note that they define the "dimension" of a real variety to be Krull dimension of its coordinate ring. I recommend reading the whole chapter.
Definition 3 is equivalent to Defintion 2 because any semialgebraic set admits a decomposition into finitely many pieces homeomorphic to $(0,1)^d$ (see Theorem 2.3.6 in RAG), and a finite union of semialgebraic sets of dimension less than $d$ cannot contain a set of dimension $d$. I.e., the only way a real variety can contain an open set homeomorphic to $(0,1)^d$ is by containing a semialgebraic set homeomorphic to $(0,1)^d$ in its semialgebraic cell decomposition.
Best Answer
(To supplement Alberto's example)
If $V$ is projective, then the gap between being locally c.i and c.i is quite big. In particular, any smooth $V$ would be locally c.i., but they are not c.i. typically. For instance, take $V$ to be a few points in $\mathbb P^2$ would give simple examples. In higher dimensions, by Grothendick-Lefschetz, if $V$ is smooth, $\dim V\geq 3$, and $V$ is c.i. then $\text{Pic}(V)=\mathbb Z$, so it is a serious restriction.
The affine case is more subtle. Again one can look at smooth varieties. If $V$ is a smooth affine curve and c.i., then the canonical bundle of $V$ is trivial. So it gives the following strategy: start with a projective curve $X$ of genus at least $2$, removing some general points to obtain an (still smooth) affine curve with non-trivial canonical bundle.
For more details on the second paragraph, see this question, especially Bjorn Poonen's comments. This paper contains relevant references, and also an example with trivial canonical bundle.