inverse element in a field of sets A field is an abelian group under addition and a group under multiplication. But for the power set of a set $S$, under union and intersection operations and with $S$ and the empty set being their identities, the inverse of any set is not unique for either union or intersection. So is the power set a field of sets?
Must the inverse of a set for either union or intersection be restricted to the complement of the set?
Thanks!
 A: A "field of sets" is a collection $\mathcal{F}$ of subsets of a given set $X$ which is closed under binary unions, intersections, and under complementation. Alas, it is not an "field" in the sense of abstract algebra.
It is a "subalgebra" of the Boolean algebra of the power set of $X$. Alas, here as well, "Boolean algebra" and "subalgebra" do not have the meaning they have in abstract algebra/ring theory: a "Boolean algebra" is not really an algebra: rather, it is a special kind of lattice.
Namely, a Boolean algebra is a set $A$ together with two binary operations, $\wedge$ (meet) and $\vee$ (join), one unary operation ${}^c$ (complementation), and two nulary operations (distinguished elements) $0$ and $1$. We require $\vee$ and $\wedge$ to be associative, commutative, and to distribute over each other (this is already different from the case of "algebra" in the ring-theoretic sense, in which only multiplication distributes over sums, and not the other way around); we also have two rules of "absorption" that further describe how $\wedge$ and $\vee$ interact: $x\wedge(x\vee y) = x$ and $x\vee(x\wedge y) = x$. Finally, we require $x\vee x^{c}=1$ and $x\wedge x^{c}=0$. 
The typical example of a Boolean algebra/Boolean lattice is the lattice of subsets of a given set $X$, with $\wedge$ corresponding to intersection, $\vee$ to union, ${}^c$ to complementation, $0$ to $\emptyset$, and $1$ to $X$. Another example is propositional calculus, with $\wedge$ corresponding to conjunction, $\vee$ to disjunction, ${}^{c}$ to negation, $1$ to the class of tautologies, and $0$ to the class of contradictions.
The reason for calling it algebra, in defiance of the meaning of "algebra" from ring theory, is historical: Boole talked about "algebra of thought" or "algebra of logic". 
Now, any Boolean algebra gives rise to a ring. We define $+$ by "symmetric difference": $x+y = (x\wedge y^{c})\vee(x^{c}\wedge y)$. If you do this, then you get an abelian group, with identity element being $0$, and with every element being its own inverse (that is, every nontrivial element is of additive order $2$). We define $*$ by $x*y=x\wedge y$. The unity of this ring is the $1$ from the Boolean lattice. This ring is an algebra over the field of $2$ elements, since it has characteristic $2$ and a unity.
You may ask: will this ring be a field? The only time it is a field is if the original Boolean algebra was the trivial algebra, with $0$ and $1$ the only elements. For if $x$ is any other element, $0\neq x$, $x\neq 1$, then if you look at the order induced by the lattice structure ($x\preceq y$ if and only if $x\wedge y = x$) then you have $0\prec x \prec 1$, and $x\wedge y \preceq x$ for all $y$. Therefore, since $x$ is strictly smaller than $1$, then so will $x*y$ for any $y$, so $x$ cannot have a multiplicative inverse in the corresponding ring.
A: One word, two meanings. A 'field of sets' is not a field in the abstract algebra sense.
EDIT: While a 'field of sets' is not a field, a Boolean Algebra (BA) of sets (of which a field of sets is a special case) is indeed an algebra. One can check that a BA is a ring, with symmetric difference $(A \cap B^c) \cup (A^c \cap B)$ as addition and intersection as multiplication. It is then trivially an algebra over the two-element field $\mathbb{F}_2$ by defining $1\cdot A=A$ and $0\cdot A=\emptyset$ and checking that $A+A=\emptyset$.
[Arturo's answer discussed this before me in a more general setting. The edit above is to set right my incorrect assertion in the comments that a BA is not an algebra.]
