We are continuing the series on non-unique factorisation. For a handy table of contents, visit the Post Series directory.

In Part 1 of this series, we introduced for a commutative ring three types of relations:

- Associaties: $a\sim b$ means that $(a) = (b)$
- Strong associates: $a\approx b$ means that $a = ub$ for $u\in U(R)$
- Very strong associates: $a\cong B$ means that $a\sim b$ and either $a = b=0$ or $a = rb$ implies that $r\in U(R)$

Here, $U(R)$ denotes the group of units of $R$. We have already seen in a ring with nontrivial idempotents like $\Z\times \Z$, a nontrivial idempotent $e$ will be satisfy $e\sim e$ and $e\approx e$, but $e\not\cong e$ because $e = ee$ and yet $e$ is not a unit and nonzero.

Therefore, $\cong$ is not an equivalence relation for all commutative rings. But it is symmetric:

*Proof.*Suppose $a\cong b$. Then $a\sim b$ and so $b\sim a$. If $a$ and $b$ are not both zero, write $a = sb, b = ta$. If $b = ra$ then $a = sra = s^2rb$. Since $a\cong b$, this implies that $s^2r$ is a unit and so $r$ is a unit. Hence $b\cong a$.

Guess what? The relation $\cong$ is also *transitive*. Since the proof is similarly short I'll leave the proof to the reader. So, $\cong$ is just missing being reflexive for all rings to be an equivalence relation for all rings. If $\cong$ is an equivalence relation for a ring $R$, then we say that $R$ is **presimplifiable**. We introduced this type of ring last time.

It's easy to see that if $\cong$ is an equivalence relation (equivalently, if it is reflexive) for a ring $R$, then $a\sim b$ in $R$ implies that $a\cong b$. Because $a\sim b\rightarrow a\approx b\rightarrow a\cong b$ in general, this means that presimplifiable rings are exactly those rings for which the concept of associates, strong associates, and very strong associates all coincide.

In the next post, we look at some examples of commutative rings that are presimplifiable, and some that are not.