Boolean Ring has $2^N$ elements "proof"

Question

I was thinking about a proof that all finite booleans rings has $2^N$ elements. My attempt was the following:

Consider a boolean ring and all of its isolated elements, i.e., elements that isn't a sum or product of other elements and are differents of zero.

For example, consider a boolean ring $R$ with isolated elements $(a,b,c,...,m)$. So, it will be also elements of the ring $a+b,a+c,..., b+c,...,a+b+c+...+m$. In fact, we can see it as the sum $\binom{m}{0}+\binom{m}{1}+...+\binom{m}{m}$ for all sums, (note that the element $0$ is covered by $\binom{m}{0}$, and every single element by $\binom{m}{1}$). So, by newton binomial the number of elements of the boolean ring will be $2^m$.

I know that is problems with this "Proof". For instance, I didn't the same thing for the products. But that is some clarity in my solutions? Indeed, I know every product times a sum will be zero, so I think it's quite fair suppose that every product is equal to some sum.

Thanks for attention.

Make sure you're not double-counting. For example, is it possible $a+b+c = b+d$? — aschepler, May 28 '24 at 03:27
it implies $a+c=d$, and by my definition, $d\neq a+c$ @aschepler — Carinha logo ali, May 28 '24 at 03:48
Exactly. It's worth pointing out why your $2^m$ sums are all different ring elements. It's not clear exactly what your "isolated elements" means, though. What is or is not a set of isolated elements of $\mathbb{Z}_2[x] / (x^3+x+1)$? — aschepler, May 28 '24 at 04:06
@aschepler are elements that aren't the the sum of another two elements — Carinha logo ali, May 28 '24 at 04:28

score 5 · Accepted Answer · answered May 28 '24 at 11:38

It isn't clear that your notion of "isolated elements" says anything.

Let's say you had an "isolated element $x$." Then for any nonzero $y$ not equal to $x$, you'd have $(x+y)+y=x$. In particular, both $x+y$ and $y$ are different from zero and add up to your isolated element.

On the other hand, this is very reminiscent of the topic of bases from linear algebra. In fact, every boolean ring is an $F_2$ algebra where $F_2$ is the field of two elements. From this it's clear that any finite boolean ring has cardinality $2^n$ where $n$ is its $F_2$ dimension.

This is similar to your idea, but in this case there's no distinguished set of "isolated elements," because anything could be part of a basis.

There is actually something that is close to what you mean, but it is slightly more involved. One can define an order on the elements by $e\leq f$ iff $ef=e$ and then deduce there are minimal nonzero elements, and that these constitute a basis. However, it's not clear that anything was bought with that argument: you may as well just have argued using a basis.

rschwieb has also given a fine Answer to this similar Question, self-contained apart from showing that Boolean rings (defined as all elements are idempotents, $x^2=x$) are commutative. The notion there of factoring the ring into smaller ones, eventually into copies of $\mathbb Z/2\mathbb Z$, gives the desired result without identifying "isolated elements". — hardmath, May 28 '24 at 13:58

Boolean Ring has $2^N$ elements "proof"

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