Let $\Omega$ be any set and let $\mathcal A$ be an algebra of sets in $\Omega$. An element $E\in \mathcal A$ is said to be an atom if there is no non-empty element $A\in \mathcal A$ such that $A\subsetneq E$. (By convention we allow $\emptyset$ also to be an atom.)
(My definition of an atom is different from the standard definition of an atom as used in measure theory.)
Question: Can each element of $\mathcal A$ be partitioned into atoms?
If $\mathcal A$ was generated by finitely many members of $\Omega$, the answer is clearly in the affirmative.
I am unable to prove (or disprove) this in the case where $\mathcal A$ is not generated by finitely many members of $\Omega$.
Can somebidy help?
No, not necessarily.
If this is true, we say that the algebra is atomic. But not every Boolean algebra is atomic, or even has atoms.
Consider for example $\mathcal P(\Bbb N)$ as a Boolean algebra, and $\rm fin$ as the ideal of finite sets there. If you haven't seen that the quotient of a Boolean algebra by an ideal is a Boolean algebra, this is a good exercise. Now $\cal B=P(\Bbb N)/\rm fin$ is a Boolean algebra, and $[A]\leq_\mathcal B[A']$ if and only if $A\setminus A'$ is a finite set.
So given any $[A]$ in $B$, if it is non-zero then $A$ is infinite, and we can split $A$ into two infinite subsets, $A_0$ and $A_1$, in which case $[A_0]<_\mathcal B[A]$.
But wait, you say, this is a general Boolean algebra! And the question is about a field of sets. But rest assured, the Stone Representation theorem tells us, amongst other things, that every Boolean algebra is isomorphic to a field of sets. And if $B$ a field of sets isomorphic to $\cal B$ above, then $B$ itself cannot have any atoms either.
(One nice thing about atomless Boolean algebras is that their first order theory is complete. To see why, you can prove via a back-and-forth argument that every two countable atomless Boolean algebras are isomorphic. This means that the theory is $\aleph_0$ categorical which in turns implies completeness.)