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.

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