Is this a correct equivalent formulation of the universal property of the polynomial ring?

To the best of my understanding, the universal property of the polynomial ring can be phrased (see page 5 here):

Given a commutative and unital $$R$$, then the polynomial ring $$R[X]$$ satisfies the universal property that for any pair $$(\phi, s)$$, where $$\phi: R \to S$$ is a ring homomorphism, $$s \in S$$, and $$S$$ is also a commutative unital ring, there is a unique ring homomorphism $$\Phi: R[X] \to S$$ such that $$\Phi(x)=s$$ and $$\Phi \circ \iota = \phi$$, where $$\iota: R \to R[X]$$ is the standard inclusion.

Note that I assume ring homomorphisms preserve the multiplicative identity.

Question: Is the following a correct equivalent formulation?

Given a commutative unital ring $$R$$, the polynomial ring $$R[X]$$ satisfies the universal property that for any pair $$(\phi, s)$$, where $$\phi: R \to S$$ is a ring homomorphism and $$s \in S$$ is such that $$\sigma \cdot s = s \cdot \sigma$$ for all $$\sigma \in \operatorname{Im}(\phi)$$, and $$S$$ is a unital ring (not necessarily commutative), then there is a unique ring homomorphism $$\Phi: R[X] \to S$$ such that $$\Phi(x) =s$$ and $$\Phi \circ \iota = \phi$$, where $$\iota: R \to R[X]$$ is the standard inclusion.

Motivation: The one-to-one correspondence between (i) $$R[X]$$ module structures on a given abelian group $$M$$ compatible with a given $$R$$-module structure on $$M$$ and (ii) (abelian) group endomorphisms on $$M$$ which are $$R$$-linear with respect to the given $$R$$-module structure, would be a direct consequence of the second formulation. (Compare page 73 here.)

Take $$S = \operatorname{End}(M)$$, $$\phi$$ the $$R$$-scalar multiplication, $$\Phi$$ the $$R[X]$$-scalar multiplication, and $$s$$ to be the $$R$$-linear (w.r.t. $$\phi$$) endomorphism of $$M$$. ($$s$$ commuting with all $$\sigma \in \operatorname{Im}(\phi)$$ is what makes it $$R$$-linear.)

Proof Attempt: I'll skip details for brevity, but basically I think I was able to show that an $$s \in S$$ commutes with all $$\sigma \in \operatorname{Im}(\phi)$$ if and only if $$s$$ is contained in some commutative subring of $$S$$ which also contains $$\operatorname{Im}(\phi)$$.

It should hopefully also be true that every such subring contains the subring generated by $$s$$ and $$\operatorname{Im}(\phi)$$ (called $$k[f]$$ on p.73 of the linked notes), and so the $$\Phi$$ we get by applying the standard formulation of the universal property of the polynomial ring should be the same regardless of which commutative subring of $$S$$ containing both $$s$$ and $$\operatorname{Im}(\phi)$$ we choose.

The other direction is trivial since obviously any $$s \in S$$ will satisfy the required condition when $$S$$ is itself commutative.

Yes, this works. The proof is basically identical to the proof for commutative rings, since the assumption that $$\sigma s=s\sigma$$ for all $$\sigma \in \operatorname{Im}(\phi)$$ means that all the elements of $$S$$ you will ever write down in the proof commute. Or alternatively, as you mentioned, you can just say that the subring generated by $$s$$ and $$\operatorname{Im}(\phi)$$ is commutative and then apply the result for commutative rings.
• Thank you for the sanity check! The notes I was looking at didn't mention this explicitly, so I just wanted to make sure I wasn't making up something that was too good to be true. It took me a while to figure out why one couldn't apply the universal property to an arbitrary element of $End(M)$ (which leads to a contradiction using additive but not linear functions) before remembering that implicit in the typical statement of the universal property is that $S$ must be commutative. This seemed like the best way to resolve this, but again I tend not to trust my own work. Commented Aug 16, 2019 at 19:07