Which complete lattices are isomorphic to a product of irreducible lattices? Given any family of complete latices $\{\mathcal{L}_i\}_{i\in I}$ s.t. for all $i\in I$ we denote $\mathcal{L}_i=(X_i,\leq_i,\wedge^i,\lor^i)$ and $X=\prod_{i\in I}X_i$ note we can define a complete lattice $\mathcal{L}=\prod_{i\in I}\mathcal{L}_i$ (call it their product) on $X$ s.t. $\mathcal{L}=(X,\leq,\wedge,\lor)$, defined for $a,b\in X$ as follows: $a\leq b\iff \forall i\in I(\pi_i(a)\leq_i\pi_i(b))$ also if $S\subseteq X$ then $\small\bigwedge_{f\in S}f=\{(i,\bigwedge^i_{f\in S}\pi_i(f)):i\in I\}$ and $\small\bigvee_{f\in S}f=\{(i,\bigvee^i_{f\in S}\pi_i(f)):i\in I\}$ in addition we call any lattice with one element trivial and say a complete lattice $\mathfrak{L}$ is irreducible if there does not exist a family of two or more non-trivial complete lattices $\{\mathfrak{L}_{i}\}_{i\in I}$ s.t. $\mathfrak{L}\cong \prod_{i\in I}\mathfrak{L}_i$. Now with all of that said, my question is when are complete lattices isomorphic to a product of irreducible lattices? For example are there any 'elementary' or 'useful' criteria for determining this? What are examples of complete lattices which are not isomorphic to any product of irreducible lattices? Could someone give me a few of these?
Obliviously any finite complete lattice is isomorphic to a product of irreducible lattices, since if the lattice itself is irreducible we are done otherwise we can factor this out into two lattices which are sublattices of the parent and thus expressible as lattices on sets each smaller then the parent set, thus repeating this process over and over eventually will provide us a family of irreducible lattices whose product is equal to our parent (this process must terminate for each of these lattices will be on smaller sized sets and by definition any trivial lattice is irreducible so if we happen to reduce any such lattice to a set on one element we're done).
In addition if any complete lattice $L_1\cong L_2\times L_3$ is not isomorphic to a prdouct of irreducible lattices  then $L_2$ or $L_3$ are not isomorphic to a product of irreducible lattices thus by applying the previous process we see any lattice not isomorphic to a prdouct of irreducible lattices must contain an infinite number of sublattices also not isomorphic to a product of irreducible lattices..
 A: For distributive lattices, there is a fairly simple way of understanding these questions.  Namely, note that if $L=A\times B$ is a product of two lattices, the elements $(1,0)$ and $(0,1)$ are complements of each other (their join is $1$ and their meet is $0$).  Conversely, if $L$ is a distributive lattice and $a,b\in L$ are complements of each other, then $L\cong A\times B$ where $A=\{x\in L:x\leq a\}$ and $B=\{x\in L:x\leq b\}$.  Indeed, there is an order-preserving map $f:L\to A\times B$ mapping $x$ to $(x\wedge a,x\wedge b)$ and the map $A\times B\to L$ sending $(x,y)$ to $x\vee y$ is inverse to $f$ since $L$ is distributive.
So, a distributive lattice is irreducible iff it has no nontrivial complemented elements.  The set of complemented elements in any distributive lattice $L$ forms a Boolean algebra which I will call $B(L)$.  Moreover, if a distributive lattice $L$ is a product $\prod_{i\in I} L_i$, then $B(L)= \prod_{i\in I} B(L_i)$.
In particular, if $L$ is a product of (nontrivial) irreducible lattices $\prod_{i\in I} L_i$, then $B(L)=\prod_{i\in I}B(L_i)\cong \mathcal{P}(I)$, since each $B(L_i)$ is just the two element lattice $\{0,1\}$.  Moreover, $L_i\cong\{x\in L:x\leq e_i\}$ where $e_i\in L$ is $1$ on the $i$th coordinate and $0$ on the others, and these elements $e_i$ are just the atoms of the Boolean algebra $B(L)$.  With this identification, the projection $L\to L_i$ is just the map $x\mapsto x\wedge e_i$.
Thus, we conclude that a distributive lattice $L$ is isomorphic to a product of irreducible lattices iff the map $f:L\to\prod_{i\in I}L_i$ is an isomorphism, where $I$ is the set of atoms of $B(L)$, $L_i=\{x\in L:x\leq i\}$, and the $i$th coordinate of $f$ is the map $x\mapsto x\wedge i$.  If $L$ is complete, these $L_i$ will automatically also be complete.  In particular, a necessary condition for $L$ to be isomorphic to a product of irreducible lattices is for $B(L)$ to be isomorphic to a power set Boolean algebra.
So, for instance, if $L$ is a complete Boolean algebra that is not isomorphic to a power set, then $L$ is not a product of irreducible lattices.  For an explicit example, $L$ could be the lattice of regular open subsets of $\mathbb{R}$, or the lattice of Borel subsets of $\mathbb{R}$ modulo sets of Lebesgue measure $0$.  For a different sort of example, $L$ could be the lattice of open subsets of the Cantor set.  Then $B(L)$ is the Boolean algebra of clopen subsets of the Cantor set, which is atomless (and in fact is not even complete).
For an example where $B(L)$ is a power set but $L$ is still not a product of irreducible lattices, you could take $L$ to be the lattice of open subsets of $\beta\mathbb{N}$.  Then $B(L)\cong\mathcal{P}(\mathbb{N})$, but its atoms are the singletons $\{n\}$ for $n\in\mathbb{N}$ so the map $L\to\prod_{i\in I}L_i$ as described above is the map $L\to\mathcal{P}(\mathbb{N})$ sending an open subset of $\beta\mathbb{N}$ to its intersection with $\mathbb{N}$, which is not injective.
