Can we describe multiplication on $\mathbb{F}_{2^n}$ as action on subsets of $n$-element set?

Question

The symmetric difference between two set $A$ and $B$ denoted $A \triangle B$ is defined as the set $(A - B) \cup (B - A)$ or equivalently $(A \cup B) - (A \cap B)$. Some years ago I was quite excited to find that the set $\mathcal{P}(N)$ of all $2^n$ subsets of an $n$-element set $N$ forms a abelian group under the $\triangle$-operation.

Being older and wiser it is easy to see where this group comes from: replacing every subset $A$ of $N$ with its indicator function, that is: by a vector of $n$ zeroes and ones indicating whether or not the first element of $N$ is in $A$, whether or not the second element is etc, we see that taking the symmetric difference of sets corresponds to adding the indicator vectors mod 2. In other words: the group we find is the underlying abelian group of the vector space $\mathbb{F}_2^n$.

Now for the question. $\mathbb{F}_2^n$ is more commonly written $\mathbb{F}_{2^n}$ and the reason is that on top of being a vector space it is also a field. In other words: there is a, really beautiful, way of mutliplying these vectors. The question then is: what does this multiplication look like when described in terms of sets? Starting from the 'abstract' field $\mathbb{F}_{2^n}$ there seem to be many ways and a lot of arbitrary choices involved when picking a way to write the elements as vectors, but perhaps, starting from sets with symmetric differences as addition there is a 'most beautiful way' to add multiplication to the picture?

What we want is a second operations on subsets of $N$ that distributes over $\triangle$ and of course an example is not hard to find: intersection will do. But looking at the vector picture again we that this gives the stupid pointwise multiplication of vectors. This is definitely not the field multiplication as it produces zero (the empty set) whenever we try to 'multiply' two disjoint subsets.

So my question is: what 'even more' beautiful set operation corresponds to the multiplication in $\mathbb{F}_{2^n}$ when letting it act on all subsets of a set of $n$ elements?

EDIT: as requested in the comments I will say something about the small cases. First for $n = 1$ we note that in this very small case pointwise and field multiplication coincide, so we can use the intersection. Thinking about intersections we note that (if nothing else) the idea of taking the field element 1 to be the full set $N$ might be something to take with us from this example to the larger cases with fancier set operations. After all, 1 is the only element besides 0 we know that is present and easily recognizable in every field, and similarly the full set is the only subset besides the empty set which is equally easy to identify in all cases.

Running with $1 = N$ leaves us with pretty much only one possibility in the the $n = 2$ case: any singleton set squared gives the other singleton set while their product equals 1 (i.e. all of $N$). Writing $c: \mathcal{P}(N) \to \mathcal{P}(N)$ for the complement-operator this 'multiplication' can be written as

$$A * B = c^{|A||B|}(A \cap B).$$ this formula looks pretty nice, but already breaks down in the $n = 1$ case. If we take the more agnostic stand point that $A * B = c^{f(A, B)}(A \cap B)$ for 'some' function $f$ we still get nowhere as this formula is bound to give elements satisfying $x^3 = 1$ while we want all elements to satisfy $x^{2^n - 1} = 1$, with at least one $x$ not satisfying $x^k = 1$ for any $k < 2^n -1$.

However, if there is any good idea in this $n = 2$ case to take with us to higher cases (similar to the idea of taking 1 to be the full set from the n = 1 case) then my bet would be it is the idea of taking $A^{-1}$ to be the complement of $A$. UPDATE: nope. Taking $A^{-1}$ to be the complement of $A$, (that is: $1 + A$ in the field view) for all elements $A$ leads to a contradiction in the $n = 3$ case. We have to think of something else.

Answer 1

I don't think you will get anything interesting from this point of view (though I could be wrong). The field $\mathbb{F}_{2^{n}}$ is very different from a boolean ring (which is the more natural setting for what you are thinking of).

One big problem is that, in your scenario (interpreting finite field elements as sets), for any any two nonempty sets $A$ and $B$, there is a set $X$ for which $A \cdot X = B$. Furthermore, the multiplicative group of a finite field is cyclic, so there would be some set $\alpha$ such that $A \cdot \alpha^{k}$ would range over all of the nonempty sets as $k$ took values from $0$ to $q-2$.

I cannot think of any intuitive set operation that would match these properties.

(Also, I would not say that $\mathbb{F}_{2}^{n}$ is more commonly referred to as $\mathbb{F}_{2^{n}}$, though it can very naturally be viewed this way. The reason to think of these objects being distinct is that the vector space $\mathbb{F}_{2}^{n}$ comes up in a lot of settings where the field multiplication doesn't make sense or is not useful to the task at hand. So I would not say that $\mathbb{F}_{2}^{n}$ is a field as well as a vector space; though it is true that with a suitably defined multiplication it can be made into a field. It can also be made into a boolean ring by defining multiplication to be set intersection.)

Answer 2

I do not know about beautiful ways to do this. However, there are no natural ways to do this, at least for $n > 2$.

Proof. Suppose there is a natural way to equip, for a finite set $X$, the set $\mathcal P(X)$ with a field structure. At the very least, naturality means that for every bijection $X \to X$, the induced map $\mathcal P(X) \to \mathcal P(X)$ is an isomorphism.

However, there are $n!$ bijections $X \to X$, but there are only $n$ automorphisms of the field $\mathbb F_{2^n}$. Since $n! > n$ for $n > 2$, we conclude that there is no natural field structure on the set $\mathcal P(X)$.

Remark. In a different direction, say that a binary operation $\odot \colon \mathcal P(X) \times \mathcal P(X) \to \mathcal P(X)$ is given by set operations if it can be defined by a combination of $\cup$, $\cap$, and $(-)^c$.

Claim. The only binary operations with identity that are given by set operations are $\triangle$ and $\cup$, and their 'complements' $(-)^c \circ \triangle$ and $\cap$. In particular, there can never be elements of order $> 2$.

Proof. Indeed, if the operation $\odot$ is given by set operations, and if $U,V \subseteq X$ are subsets set, then the validity of the statement $x \in U \odot V$ is given by some truth table $$\begin{array}{ccc}x \in U & x \in V & x \in U \odot V \\ 0 & 0 & a_{00}\\ 0 & 1 & a_{01}\\ 1 & 0 & a_{10}\\ 1 & 1 & a_{11}.\\\end{array}$$ Now suppose wlog that the identity of $\odot$ is given by $\varnothing$ (the only other alternative, by naturality, being $X$). Then $a_{00} = 0$, $a_{01} = 1$, and $a_{10} = 1$. Then either $a_{11} = 0$, so $\odot = \triangle$, or $a_{11} = 1$ and $\odot = \cup$. In the first case. every element of $\mathcal P(X)$ has order $2$, and in the second case only $\varnothing$ is invertible. The case where the identity is $X$ gives the other two possibilities.

Thus, we see for $n \geq 2$ that the field operation on $\mathbb F_{2^n}$ cannot be given by set operations, because there is an element of multiplicative order $2^n - 1$.

Remark. However, for $n=2$, if we allow the swap operator $s^* \colon \mathcal P(X) \to \mathcal P(X)$ given by pulling back along the transposition $s \colon X \to X$, we can describe the field structure: one can show by hand that the operation $$U \cdot V = (U \cap V) \triangle ((s^*U \triangle U) \cap (s^*V \triangle V))$$ gives the field operation on $\mathbb F_4 = \mathbb F_2[\alpha]$, where $\alpha^2 = \alpha + 1$. Here, $\alpha$ corresponds to one of the singleton sets in $X$, and $\alpha^2$ to the other one (of course, by naturality it shouldn't matter which is which!), whereas $0 = \varnothing$ and $1 = X$. Using distributivity of $\cap$ over $\triangle$, we can rewrite this as $$U \cdot V = (s^*U \cap s^*V) \triangle (s^*U \cap V) \triangle (U \cap s^*V).$$ One shows that the operations agree $\triangle$ and $\cdot$ agree with addition and multiplication on $\mathbb F_4$ respectively, and hence a fortiori the product $\cdot$ is associative, etc.

Finally, in the case $n=1$, the answer is trivially yes: the usual operations ($\triangle$ and $\cap$) give the field of $2$ elements.

Remark. We can use similar methods to show that there are no natural operations on $\mathcal P(X)$ making it isomorphisc as a ring to $\mathbb Z/2^n \mathbb Z$: the latter has only $1$ ring automorphism.

Our second argument even shows that the underlying group structure on $\mathbb Z/2^n \mathbb Z$ is not given by set operations for $n > 1$, for it has elements of order $> 2$.