Is the extension ring still defined as polynomials evaluation when the generator doesn't commute with its elements?

Question

Given a ring $R$ and an element $a$, then $R[a]$ is defined to be the smallest ring containing both $R$ and $a.$ In Galois Theory class, I learnt that if $a$ commutes with every element in $R$, then $R[a] = \{f(a) \mid f \in R[x]\}$. However I noticed that when there exists an element $r \in R$ such that $ra \neq ar$ then I can't find such $f(x)$ such that $f(a)=ar$ even though this element belongs with the ring extension. I also learnt somewhere that you can only define a polynomial ring if the indeterminant $x$ commutes with $R$. Does this mean defining $R[a]$ as polynomials is impossible for this case? Did I get it wrong somewhere? Hope someone could point it out for me.

Answer 1

Does this mean defining $R[a]$ as polynomials is impossible for this case?

It is standard practice to define the ring of polynomials this way, with $x$ commuting with elements of $R$.

You then say “but evaluation does not work unless you are evaluating at a central element.”

Now, a twisted polynomial ring $R[x;\sigma]$ includes a ring endomorphism of $R$ to define noncommutative behavior between $R$ and $x$: $xb:=\sigma(b)x$. So people do think about such rings too.

If $a$ is in a ring containing $R$, you can still define $R[a]$ as the smallest ring containing $R$ and $a$. What you lose is a ring homomorphism from $R[x]\to R[a]$ using evaluation. You can, however, still map from the free algebra $R\langle x\rangle$. Maybe this latter thing is what you are interested in.