Posted by Jason Polak on 11. April 2017 · Write a comment · Categories: commutative-algebra · Tags:

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:

  1. $a\sim b$ means that $a$ and $b$ are associates
  2. $a\approx b$ means that $a$ and $b$ are strong associates
  3. $a\cong b$ 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 $U(R)$:

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

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