If $F$ is a field then the polynomial ring $F[x]$ is a unique factorisation domain: that is, every nonunit can be written uniquely as a product of irreducible elements up to a unit multiple. So in $\Q[x]$ for example, you can be sure that the polynomial $x^2 – 2 = (x-2)(x+2)$ can't be factored any other way, and thus the only zeros of $x^2 – 2$ really are $\pm 2$.

If $F$ is not a field, then a polynomial might have a bunch of different factorisations. For example, in the ring $\Z/4[x]$ we can write $x^2 = (x)(x) = (x+2)(x+2)$. How can we make sense of factorisations in rings that are not unique factorisation domains? In order to do so, we first should make sure we understand what **irreducible** means in this context.

In the next several posts we'll look at non-unique factorisation more closely, following a paper of Anderson and Valdes-Leon [1], but keeping the posts self-contained. We'll start by looking at the concept of associates. One can in fact look at several different variations of associates. Here are three:

**Definition.**Let $R$ be a commutative ring. Two elements $a,b\in R$ are called

**associates**if $(a) = (b)$. They are called

**strong associates**if $a = ub$ for some unit $u\in R$. Finally, they are called

**very strong associates**if $a$ and $b$ are associates and either $a=b=0$ or $a=rb$ implies that $r$ is a unit of $R$.

These are three successively strong types of associates. Now, usually one would say that $a$ is irreducible if $a = bc$ implies that $b$ or $c$ is an associate of $a$. Now, one can replace associate by strong or very strong associate and then one gets three different possible definitions of irreducible. We will use the following symbols for the defined relations:

- $a\sim b$ means that $a$ and $b$ are associates
- $a\approx b$ means that $a$ and $b$ are strong associates
- $a\cong b$ means that $a$ and $b$ are very strong associates

It is easy to prove that $\sim$ and $\approx$ are both equivalence relations. On the other hand, $\cong$ is not an equivalance relation for all rings. Indeed, if $e\not\in \{0,1\}$ is an idempotent of a ring $R$ then $e\sim e$ but $e = ee$, and $e$ is not a unit! Therefore, in any ring $R$ that has nontrivial idempotents, the relation $\cong$ of being very strong associates is not reflexive. Because $\sim$ and $\approx$ are both equivalence relations, any ring with nontrivial idempotents like $\Z\times \Z$ shows that the concept of very strong associates is indeed stronger than strong associates.

**Definition.**We will call a commutative ring

**presimplifiable**if $\cong$ is reflexive on $R$.

It is easy to see that $R$ is presimplifiable if and only if for all $x=xy$ implies that $x=0$ or $y\in U(R)$ for all $x,y\in R$. Using this observation, we'll end this post with another equivalent condition for a commutative ring to be presimplifiable.

**Theorem.**The following conditions are equivalent for a commutative ring $R$ with group of units $U(R)$:

- $R$ is presimplifiable.
- $Z(R)\subseteq 1 – U(R)$.

*Proof.*$1\Rightarrow 2$. Suppose $R$ is presimplifiable. Let $x\in Z(R)$ and $y\in R$ be nonzero and such that $xy=0$. Then $y(1-x) = y$ and so since $1-x\not=0$ we must have $1-x\in U(R)$ and hence $x\in 1 – U(R)$.

$2\Rightarrow 1$. Suppose that $Z(R)\subseteq 1 – U(R)$. Let $x\in R$ and suppose that $x = xy$. If $x\not=0$ then $x(1 – y) = 0$ and thus $1- y\in 1 – U(R)$. Hence $y\in UR(R)$. QED

In the upcoming post, we'll look more closely at some of the properties of these three different types of relations we've introduced.

[1] Anderson, D. D.; Valdes-Leon, Silvia. Factorization in commutative rings with zero divisors. Rocky Mountain J. Math. 26 (1996), no. 2, 439–480