# Quotient of a monoid by equivalence relation

In abstract algebra (ring theory specifically), when we are dealing with factorization, UFD's, and so on, we are often only interested in the multiplicative structure of the ring, not the additive structure. So here is the basic situation we face: we a start with an integral domain $(R,+,*)$ (i.e. a commutative ring with no zero divisors). Then from that we take the multiplicative monoid $(M,*)$ where $M = R \setminus \{0\}$. The monoid $M$ is commutative and has the cancellation property: $ab=ac$ implies $b=c$. (A monoid is a set endowed with an operation that is associative and has an identity.)

In the context of factorization and associated topics, we also don't care about multiplication by units (invertible elements). We call two elements $a$ and $b$ of $M$ "associates" if $a=bu$ for some unit $u$. So my question is, can we usefully construct out of $M$ yet another structure which consists of equivalence classes under the associate equivalence relation? You can take a quotient of a group with a subset if the subset is a normal subgroup. You can take a quotient of a ring with a subset if the subset is an ideal. So what is the condition that a subset of a monoid has to satisfy in order to be able to construct a quotient monoid? And whatever the condition is, does the group $G$ of units of $M$ satisfy this condition, so that $M/G$ would be our quotient monoid, and would be the quotient of M under the associate equivalence relation? And assuming that we are allowed to form the quotient monoid $M/G$, what is the structure of this monoid?

Finally, I would like to do this for the particular example where $M$ is the monoid of $2$ by $2$ matrices with integer entries (note : it is not commutative) and $G=\mathrm {GL_2}(\Bbb Z)$ or $G=\mathrm {SL_2}(\Bbb Z)$.

• Your example is unclear to me. In general the multiplicative monoid $M$ of an integral domain will have a subset $U$ which is a group, and you'll look at $U \setminus M$ the set of orbits of $U$ acting on $M$. If you want to remove one more element $a$, do the same with $V$ the unit group of $a^{-1} M$. For defining $a^{-1} M$ you'll need what Daron wrote, ie. the rule of equivalence of fractions that we use for constructing the quotient field. – reuns Jul 21 '17 at 22:39

You need the equivalence relation to be a congruence.

The equivaence relation $\sim$ on the monoid $M$ is called a congruence to mean that

$(x \sim a$ and $y \sim b)\implies xy \sim ab$ and $yx \sim ba$ for all $x,y,a,b \in M$

In this case we form the group of equivalence classes and define multiplication just like for groups: We define $[x][y] = [xy]$. The relation being a congruence ensures this is well-defined.

When $M$ is a group we can prove $[1]$ is a normal subgroup, and the cosets of $[1]$ are exactly the partition elements. For monoids it's still true that $[1]$ is closed under inverses whenever they exist.

That's because $x \in [1] \implies x \sim 1 \implies x^{-1}x \sim x^{-1} \implies 1 \sim x^{-1} \implies x^{-1} \in [1]$.

But in general $[1]$ is not a normal subgroup; and the cosets don't correspond to the partition elements; and don't even form a partition!

For example consider $(\mathbb N,\times)$ under the parity relation. It's easy to see $[1] = \{1,3,5, \ldots\}$ is just the odd numbers. We might like to have the coset $2[1] =$ the even numbers. But unfortunately $2[1] = \{2,6,10, \ldots\}$. Moreover $[2] = \{2,4,6, \ldots\}$ is different to $2[1]$.

In your particular case you want to define $(x \sim y \iff x=uy$ for some invertible $u)$. This is indeed a congruence so we can define the quotient ring as above. Moreover the partition elements turn out to be what we want: $[1]$ is the normal subgroup $U$ of all units. To get the class of $x$ just take all left-multiples by units. In other words $[x] = Ux = [1]x$.

We could replace $U$ with any normal subgroup of the group of units and this would work.

• An alternate definition of congruence would be: an equivalence relation on a monoid $M$ is a congruence if and only if it is also a submonoid of $M \times M$. (It should be trivial to verify the equivalence.) – Daniel Schepler Jul 21 '17 at 22:50
• Important to note that your relation $x \sim y \iff x = uy$ is only a congruence relation because M is commutative! Otherwise you need to generate a slightly more complicated equivalence relation here. – Morgan Rogers Nov 11 '18 at 13:47