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Let $[a]$ be the equivalence class of associates of an element $a$ of a ring/rng.

Two elements of a ring/rng are associates ($\sim$) if they generate the same principal ideal.

$[a] \cdot [b] \sim [a \cdot b]$ in any commutative rng $R$ in the following way:

  1. For any element $a'$ from $[a]$ and $b'$ from $[b]$: $a' \cdot b' \sim c$ for some $c$ from $[a \cdot b]$;
  2. For any element $c$ from $[a \cdot b]$: $c \sim a' \cdot b'$, for some $a'$ from $[a]$ and $b'$ from $[b]$.

Proof:

  1. If $a' \sim a$, then $a' = r \cdot a + na$ for some $r$ from $R$ and an integer $n$;
    If $b' \sim b$, then $b' = s \cdot b + mb$ for some $s$ from $R$ and an integer $m$; $a' \cdot b' = (r \cdot a + na) \cdot (s \cdot b + mb) = (r \cdot s + ns + mr) \cdot (a \cdot b) + nm(a \cdot b)$;
    Thus, $a' \cdot b'$ is in the principal ideal of $a \cdot b$;
    Exchanging $a'$ with $a$, and $b'$ with $b$ in the formulas above:
    $a \cdot b$ is in the principal ideal of $a' \cdot b'$;
    Therefore, $a' \cdot b' \sim a \cdot b$, where $a \cdot b$ is an element of $[a \cdot b]$.

  2. $c \sim a \cdot b$ for any element $c$ from $[a \cdot b]$ by the definition of the class of associates.

I am wondering in which types of rings/rngs we can write $[a] \cdot [b] = [a \cdot b]$ meaning:

  1. For any element $a'$ from $[a]$ and $b'$ from $[b]$: $a' \cdot b' = c$ for some $c$ from $[a \cdot b]$;
  2. For any element $c$ from $[a \cdot b]$: $c = a' \cdot b'$, for some $a'$ from $[a]$ and $b'$ from $[b]$.

The first part of the definition is identical to the one for $[a] \cdot [b] \sim [a \cdot b]$
since $a' \cdot b' \sim a \cdot b$ means $a' \cdot b'$ is an element of $[a \cdot b]$.

I need help with the second part of the definition.
The widest class of rings where $c \sim a \cdot b$ means $c = a' \cdot b'$ for some $a' \sim a$ and $b' \sim b$
I found so far is integral domains:

If $c \sim a \cdot b$ in an integral domain, then $c = u \cdot (a \cdot b)$ for some unit $u$;
Then $c = (u \cdot a) \cdot b$, where $u \cdot a \sim a$ and $b \sim b$.

However, it looks like the formula $[a] \cdot [b] = [a \cdot b]$ works for all cyclic rings/rngs.
Is it possible to find a wider class than integral domains?
Is it true in commutative principal ideal rings/rngs?

Update

The formula $[a] \cdot [b] = [a \cdot b]$ works for all cyclic rngs:

  1. $[a] \cdot [b] = [a \cdot b]$ in a prime-power cyclic rng
    (since $a \sim p^k$ for any element $a$ of a prime-power cyclic rng
    https://math.stackexchange.com/a/3513734/427611);

  2. $[(a, b)] = ([a], [b])$ for the direct product of rngs
    ($(a, b) \sim (c, d) \iff a \sim c \land b \sim d$);

  3. Any finite cyclic rng is the direct product of prime-power cyclic rngs.

I am trying to apply the same logic to all commutative principal ideal rngs
using the Zariski–Samuel theorem (Structure theory for commutative PIR's):
$[a] \cdot [b] = [a \cdot b]$ in principal ideal rngs if
$[a] \cdot [b] = [a \cdot b]$ in special principal rngs.

I need help checking the formula for special principal rngs.

Alex C
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  • I am not sure that I understood you. It seems that in any commutative ring with $1$, you can define $[a] \cdot [b] =: [ab]$ unambiguously. – Sasha Feb 29 '20 at 13:24
  • @Sasha: How do we know if $[a] \cdot [b]$ gives you all elements of $[a \cdot b]$? – Alex C Feb 29 '20 at 13:26
  • Why you need all elements? This is an equivalence class. We define $[a]\cdot[b]$ to be $[ab]$, and then need to check that if $[a^{\prime}]=[a]$ and $[b^{\prime}]=[b]$ then $[a^{\prime}b^{\prime}]=[ab]$. Indeed, $a^{\prime}=ua$ for some $u$, $b^{\prime} = vb$ for some $v$, and so $a^{\prime}b^{\prime}=uvab$. Symmetrically, $ab$ is a multiple of $a^{\prime}b^{\prime}$, and therefore $ab$ and $a^{\prime}b^{\prime}$ generate the same ideal, so $[a^{\prime}b^{\prime}]=[ab]$... No? – Sasha Feb 29 '20 at 13:29
  • Oh, I see now, that maybe you don't define $[a] \cdot [b]$ to be $[ab]$, but as the set of products, or sums of products, or whatever... I don't think that would be my approach. I would simply try to define a multiplication on the set of equivalence classes, abstracting myself from this product of sets thing. – Sasha Feb 29 '20 at 13:31
  • So I define $[a] \cdot [b]$ to be $[ab]$, by definition... – Sasha Feb 29 '20 at 13:31
  • @Sasha You are correct. I am looking for an answer for a different definition. – Alex C Feb 29 '20 at 13:34

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