euclidean division for polynomials with coefficients in noncommutative rings

Question

I know that if $R$ is a commutative unitary ring, $f(x), g(x) \in R[x]$, and the leading coefficient of $g$ is a unit in $R$, it is possible to divide $f$ by $g$. My question is: is commutativity of $R$ necessary? I have this doubt, because everywhere the theorem is stated with commutativity as hypothesis, but I'm not able to see where it is used in the proof. Thank you

Answer 1

No, it is not necessary. As you correctly observed, the usual proof goes through. You have a choice of whether to do left or right division though, i.e., whether you want $f=gq + r$ or $f=q'g +r'$.

More generally, it also works in (right) skew polynomial rings of the type $R[x;\sigma,\delta]$ where $\sigma$ is an endomorphism of $R$, $\delta$ is a $\sigma$-derivation, and multiplication is defined by $ax= x\sigma(a) + \delta(a)$ for $a \in R$. However, now you need to be somewhat careful with regards to left/right division: If you write coefficients on the right (like I do here), the right division (of the type $f=gq+r$) will work. But if $\sigma$ is not surjective, the left division may not work.

I believe most texts, unless they are explicitly concerned with noncommutive rings, simply do not treat polynomial rings over noncommutative rings at all.