Equal annihilator implies associates Let $\mathbb Z_n$ be the ring of integers modulo $n$. Let $x,y \in \mathbb Z_n$ be such that $ann(x) = ann(y)$ where $ann(x) = \{a \in \mathbb Z_n : ax =0 \text{ in } \mathbb Z_n\}$. Then $(x,n) = (y,n)$ and hence $x = u y$ for some unit $u \in \mathbb Z_n$.
is this result is true in an arbitrary finite commutative ring with unity?
Thank you.
 A: Counterexample
In $R=F_2[x,y]/(x,y)^2$, $ann_R(x)=ann_R(y)=(x,y)$ but $xR\cap yR=\{0\}$.
An extension
If $R$ in any ring with the annihilator hypothesis that you wrote for $x,y\in R$, we have, by homomorphism theorems, that the left multiplication homomorphism factors through the right multiplication homomorphism to produce a homomorphism $g:xR\to yR$ such that $g(xr)=yr$ for all $r\in R$.
Because $R$ is finite and the annihilators are equal, $|xR|=|yR|$, and since $g$ is necessarily onto, this means that $g$ is an isomorphism between $xR$ and $yR$.
If $R$ is additionally right self-injective, $g$ extends to $\tilde{g}$ from $R\to R$. At this point, $\tilde{g}(1)xr=yr$ for all $r\in R$.
We claim that $\tilde{g}(1)$ is not a zero divisor.  If it were, and $r\neq 0$ was such that $\tilde{g}(1)r=0$, we'd have that $0=\tilde{g}(1)xr=g(xr)=yr$, but this contradicts the injectivity of $g$ on $xR$.  We must conclude $\tilde{g}(1)$ is not a zero divisor.
Now in a finite ring, each element is either a unit or a zero divisor, so we must conclude that $\tilde{g}(1)$ is a unit. In the special case of $r=1$ that reads that $ux=y$ where $u=\tilde{g}(1)$.
So the theorem holds for finite commutative self-injective rings. All of the rings $\mathbb Z/(n)$ are such rings.
