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Question 1. For a commutative ring $R$, if we view $R$ as a commutative monoid, is there an algebraic structure corresponding to $R/R^{\times}$, and does it have a meaning? In detail, is $R/R^{\times}$ a commutative monoid?

Question 2. Is the following statement correct? or can we say the following?

If $R$ is a UFD, then every non-zero element $R/R^{\times}$ is a product of irreducible elements of $R/R^{\times}$, which is unique up to order.

Background.

Let $R$ be a commutative ring with multiplicative identity. Then we can view $R$ as a commutative monoid. We know that $R^{\times}$, the set of units, forms a group, called the group of units of $R$.

If we consider the group action of $R^{\times}$ on $R$ via multiplication, then two elements of $R$ are associate if and only if they are in the same $R^\times$-orbit. In other words, the ring $R$ can be partitioned into the set of $R^\times$-orbits.

Example.

  • If $R= \mathbb{Z}$, then $R^\times = \{-1,1\}$ and $R/R^\times = \{0,1,2,\ldots\}=\mathbb{N}_0$.
  • If $R=\mathbb{Z}_6$, $R^\times = \{1,5\}=\{-1,1\}$ and $R/R^{\times}=\{0,1,2,3\}$.
  • If $R=\mathbb{Z}_7$, $R^\times = \{1,2,3,4,5,6\}$ and $R/R^{\times}=\{0,1\}$.
with-forest
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1 Answers1

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If $M$ is a monoid, we can let $M^*$ be its group of units.

We can define a binary relation on $M$ by $$a\sim b\iff \exists u\in M^*\text{ such that }a=ub.$$ This is an equivalence relation:

  1. $a=1a$, so $a\sim a$ for all $a\in M$.
  2. If $a=ub$ then $b=u^{-1}a$, so $a\sim b$ implies $b\sim a$.
  3. If $a=ub$ and $b=vc$, then $a=uvc$, and since $uv\in M^*$, then $a\sim b$ and $b\sim c$ implies $a\sim c$.

When $M$ is commutative (or when $M^*$ is central), this is also a congruence: if $a\sim b$ and $c\sim d$, then $a=ub$ and $c=vd$. then $ac=ubvd = (uv)bd$, so $ac\sim bd$.

Since $\sim$ is a congruence, we can construct the quotient $M/\sim$. This is essentially the "$M/M^*$" you are describing (but in monoids we do not mod out by submonoids, we have to mod out by congruences). This quotient is always a monoid, and these quotients satisfy the Isomorphism Theorems from General Algebra.

Your second statement is correct if you take the empty product to be the identity of multiplication.

Arturo Magidin
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