[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]
Below, $R$ is a commutative ring with $1$.
In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.
The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.
I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.
So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.
My questions:
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If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution 🙂 I'll post my solution once I have one.
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In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
Best Answer
Answer to the title question: Yes, $(a) = (b)$ in $\mathbb{Z}/(n)$ iff $a, b$ are associates in $\mathbb{Z}/(n)$.
Hint: by Chinese Remainder it suffices to consider the case $n = p^k$ is a prime power. Then show that every ideal of $\mathbb{Z}/(p^k)$ is of the form $(p^n + (p^k))$ for some $n = 0, \ldots, k$. Then check directly that $p^n + (p^k)$, $p^m + (p^k)$ are associates iff $n = m$.
For a "universal" counterexample outside $\mathbb{Z}/(n)$, take $R = k[x,y]/(yx^2-y)$, where $k$ is any field: here $(y) = (xy)$, but $y$ and $xy$ are not associates.
Edit: The last point is subtle, and deserves clarification. If $\text{char}(k) \ne 2$, there is an injection
$$R \hookrightarrow k[x,y]/(y) \times k[x,y]/(x^2-1) \cong k[x] \times (k[x]/(x-1))[y] \times (k[x]/(x+1))[y]$$
(notice $(x-1), (x+1)$ are comaximal in $k[x]$ since $\text{char}(k) \ne 2$). This induces an injection
$$R^\times \hookrightarrow (k[x] \times (k[x]/(x-1))[y] \times (k[x]/(x+1))[y])^\times \cong k^\times \times k^\times \times k^\times$$
A unit in $R$ is thus (the image of) a polynomial $f \in k[x,y]$ that is simultaneously a nonzero constant modulo $y, x-1$, and $x+1$. Being a unit modulo $y$ means $f = u + yg$ for some $g \in k[x,y]$, $u \in k^\times$. Then $f \bmod (x-1)$ constant in $k[y] \implies g(1,y) = 0$ (otherwise $\deg_y f(1,y) > 0$), so $(x-1) \mid g$ (by expanding $g$ in terms of $x-1$). By the same reasoning $(x+1) \mid g$, so in fact $(x^2-1) \mid g$. Thus $f = u + y(x^2-1)h \implies f \equiv u$ in $R$ is constant.
If $\text{char}(k) = 2$, then $x^2-1 = (x+1)^2$. Setting $S := k[x,y]/(x+1)^2$, there is an injection
$$R^\times \hookrightarrow (k[x,y]/(y) \times S)^\times \cong k^\times \times S^\times$$
Under the projection $S \to k[y]$ (quotienting by $x+1$), every unit in $S^\times$ differs modulo $x+1$ from a unit in $k[y]^\times = k^\times$. Thus $S^\times = k^\times + (x+1)S$, so if $f \in k[x,y]$ reduces to a unit in $S$ and $k[x]$, $f = u + (x+1)g$ for some $u \in k^\times, g \in k[x,y]$. Moreover $y \mid g$ (otherwise $\deg_x f(x,0) > 0$), so in fact $f = u + y(x+1)h$, which are indeed all units in $R$ (since $(y(x+1))^2 = 0$ in $R$, so $f^2 = u^2$). Thus $R^\times = \{k^\times + y(x+1)h \mid h \in R\}$, so for any $v \in R^\times$, $xy \ne vy$ (indeed, either $vy \in k^\times y$ or $\deg_y(vy) \ge 2$).