Given a commutative ring with unity $R$, its units $U(R)$ form a multiplicative group.

$U(R)$ acts on $R$ by multiplication. The orbits of the action are called associates, and the resulting equivalence relation is called associatedness.

For example, $ℤ$ has units $\{-1,1\}$, and the associatedness is $\{\{-n,n\}:n\inℤ_{≥0}\}$.

What happens if we drop the commutativity of $R$? Since left and right multiplication differ, $U(R)$ will act differently for each version.

I chose upper triangular matrices on $\text{GF}(2)$ for an example, and acquired the following "left associates": $$ \{\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix},\begin{pmatrix}1 & 1 \\ 0 & 1\end{pmatrix}\},\{\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}1 & 1 \\ 0 & 0\end{pmatrix}\},\{\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix},\begin{pmatrix}0 & 1 \\ 0 & 1\end{pmatrix}\} $$

The units act as elementary row operations. For "right associates," the units will act as elementary column operations instead.

This arises the following question: Will "left associates" and "right associates" ever coincide for a noncommutative $R$?

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    $\begingroup$ Your question can be rephrased as "under which conditions is $U(R)$ a normal subgroup of $R-{0}$ (besides $R$ being a commutative ring)?" $\endgroup$ – Learning Jan 30 at 1:45
  • $\begingroup$ and the title is misleading. if $R$ is not commutative, then $U(R)$ does act on $R$ with a priori two different actions: left and right. $\endgroup$ – Learning Jan 30 at 1:46
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    $\begingroup$ @BQT Generally $R-0$ is not a multiplicative group (for example, $R=ℤ$), so I don't think so. $\endgroup$ – Dannyu NDos Jan 30 at 1:47
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    $\begingroup$ You're right. Then take any non commutative ring for which $U(R)=R-0$ (e.g. $GL_n(\mathbb{R})$), then the action of $U(R)$ on $R$ has only one set of associates: the entire $R$ (plus $\{0\}$. That gives an example of non commutative ring with "left associates" being identical to "right associates". $\endgroup$ – Learning Jan 30 at 1:53
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    $\begingroup$ @BQT But in that case, what is the ring structure on $R=GL_n(\Bbb R)$? It is a multiplicative group, but it is not stable by sum (and the zero matrix doesn't belong to $R$). $\endgroup$ – Nal Jan 31 at 15:04

Take any non commutative division ring $R$, for example the quaternions.

Since every non zero element is invertible, the only orbits for left or right multiplication are $\{0\}$ and $ R- \{0\}$. Hence left and right associates coincide.


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