Remark

The ring $A[X]_{\sigma ,d}$ has the following universal property: given a ring $A'$, a ring homomorphism $f: A\rightarrow A'$, and an element $x$ of $A'$ such that $xf(a) =f(\sigma (a))x+f(d(a))$ for every $a\in A$, there exists a unique ring homomorphism $g: A[X]_{\sigma ,d}\rightarrow A'$ that extends $f$ and maps X onto $x$.

The uniqueness is clear. Let us therefore show that the mapping $g: A[X]_{\sigma ,d}\rightarrow A'$ defined by $g(\sum a_nX^n) =\sum f(a_n)x^n$ has the required properties. It extends $f$, maps X onto $x$, and is a group homomorphism. We have $g(1) = 1$. For $a\in A$ and $Q =\sum a_nX^n$ in $A[X]_{\sigma ,d}$, we have

as well as

$g$(XQ) $=g(\sum (\sigma (a_n)X^{n+1}+d(a_n)X^n))$

It follows that we have $g(P)g(Q) =g$(PQ) for $P,Q$ in $A[X]_{\sigma ,d}$ and therefore that $g$ is a ring homomorphism.