A 'rig' R has a commutative associative addition, an associative multiplication that distributes over addition, an element 0 with r+0 = r and 0r = 0 = r0 for all r ∈ R, and an element 1 with 1r = r = r1 for all r ∈ R.

A rig is 'idempotent' if rr = r for all r ∈ R.

The free idempotent rig on two generators is finite. But how many elements does it have?

It has at most 4⁷ elements, but in fact far fewer.

(1/n)

#algebra

We can start by taking two elements a and b and multiplying them in all ways subject to associativity and the idempotent law. We get just 7 elements:

1, a, b, ab, ba, aba, bab

Then we start adding these. In an idempotent rig r + r + r + r = r + r for any element r, so we get at most 4⁷ elements.

Why is that equation true? Because

(1+1)² = 1 + 1

(2/n)

But the free idempotent rig on two generators has far fewer than 4⁷ elements! After all, it obeys many more relations, like

a + ab + ba + b = a + b

(Notice that we can't get from this to ab + ba = 0, since we can't subtract.)

So, how many elements does it have?

By the way, the free idempotent rig on 3 generators is probably infinite, since there are infinitely many 'square-free words' with 3 letters:

en.wikipedia.org/wiki/Square-f

(3/n, n = 3)

@johncarlosbaez hang on do you mean a +ab +ba+b=a+b?

@johncarlosbaez
I believe the number is small enough that it's hand-countable (the figure I've gotten is 278 though it's bashy enough I've likely made a mistake)

@alexthecamel @johncarlosbaez
Taking the last few observations into account, my current upper bound count contains at least...
Terms where at most one term of each degree appears,
$$1+3 \times 2^3 \times 3^3 + (2 \times 3) \times 2^2 \times 3^2 + (2\times 3) \times 2 \times 3 + 2 \times 3 = 907$$
That's "0", terms with non-zero deg 1 coeff, terms with non-zero deg two coeff etc.
... that's already more than @alex counted, are there identities for terms like $$1 + a + ab + aba$$ I'm missing?

@rogers @johncarlosbaez
you do have ab+ba=(ab+ba)^2=ab+ba+bab+aba=(aba+bab)^2=aba+bab, right?

@alexthecamel @johncarlosbaez that specific one wouldn't affect this count, but maybe there's a similar one for $$ab + aba$$?

@rogers @johncarlosbaez
will get a picture of the categories I have when my phone charges enough

@rogers @johncarlosbaez
hopefully it is not too unclear what I mean here

@rogers @johncarlosbaez
of course 4^7 is a small enough number that you could just computationally build a graph with vertices connected with the x+y=x+xy+yx+y eq relation and find connected components

@alexthecamel @johncarlosbaez so far we have only been using two-term relations. Many of the three-or-more term relations are already accounted for that way, but there are enough possibilities that one should probably try to prove that there are no extra reductions?

@rogers @johncarlosbaez
i think >2 term relations don't add any more constraints: eg
x+y+z=x+y+xy+yx+z=x+y+z+xy+yx+xz+zx=x+y+z+xy+yx+xz+zx+zy+yz=(x+y+z)^2

@rogers @johncarlosbaez
I guess I've not accounted for relations of the form
x(y+z)^2=x(y+z)? I don't think I'm missing any which are nontrivial but that is a bit of an oversight

@rogers @johncarlosbaez
ok i've updated my code to just brute force check + and * are well defined, so there's no problems here
if i'm doing the free rig on n variables i'd have to be more subtle I assume.
github.com/agunning/freerig

a Schelling point for those who seek one