The monoid (F_n[j], *)/(a + bj ~ b + aj) is isomorphic to (F_n X F_(n+1)/2, *), when such fields exist.

Keith J. Bauer, 2024

Theorem: Let F_n be a finite field of size n and let F_m be a finite field such that 2m - 1 = n. (Note that m and n are necessarily prime powers.) Let j^2 = 1 and let ~ be the equivalence relation on the elements of F_n[j] defined by a + bj ~ b + aj (with no further unnecessary equivalences). Given these conditions, (F_n[j], *)/~ is isomorphic to (F_n X F_m, *).

First, we need to prove a lemma.

Lemma. Let R be a finite ring of size n such that 2m - 1 = n. Then there exists a ring with a multiplication monoid isomorphic to (R[j], *)/~ iff there exists a ring with a multiplication monoid isomorphic to (R, *)/-1, where -1 represents the relation x ~ -x for all x in R.

Proof. Let R be a finite ring with an odd number of elements. Then 2 is a unit in R so there exists an isomorphism f from R[j] to R^2 given by

f(a + bj) = (a + b, a - b)

which has the inverse

f^-1(a, b) = (a + b)/2 + (a - b)/2 * j.

By this isomorphism we can re-characterize the action of swapping the coefficients of an element of R[j]:

f(b + aj) = (a + b, b - a).

Thus, for R where 2 is a unit, the action is equivalent to multiplying the right element of an element of R^2 by -1. We can now write

(R[j], *)/~ is isomorphic to (R, *) X (R, *)/-1.

Now we use that R is finite: if |R| = 2m - 1, then note that all non-zero elements cannot be their own additive inverse: a = -a implies 2a = 0 and because 2 is a unit, a = 0. Therefore |(R, *)/-1| = m. Additionally, gcd(2m - 1, m) = 1 because -1 * (2m - 1) + 2 * m = 1. According to Rains in https://oeis.org/A037291/a037291.txt, finite rings may be decomposed uniquely into products of rings of prime power order. Combining all of this information, a ring with a multiplication monoid of size m(2m - 1) must be the product of a ring of size m and 2m - 1. In fact, we already have a ring of size 2m - 1, which is simply R. Therefore the problem reduces to finding a ring with a multiplication monoid isomorphic to (R, *)/-1 as desired.

Now, let R be a field of size n, with n an odd prime power. It is well-known that the unit group of finite fields are cyclic. It is also a trivial observation that (R, *) is simply the unit group of R, isomorphic to C_2m-2, with a single absorbing element added. Then (R, *)/-1 must be the same but with unit group isomorphic to C_m-1. Such a ring must have all non-zero elements invertible, thus also making it a field. The necessary and sufficient condition must be that F_m exists, that is, m is a prime power, as desired.