TITLE: Why is commutativity needed for polynomial evaluation to be a ring homomorphism?
QUESTION [6 upvotes]: In this proof they prove it for fields, but they say that the proof holds for a commutative ring with unity. But why do we need it to be commutative, and why do we need a unity? Why does it not hold if we are only working with a ring?
In the last part of the proof they work with $1$, is this why we need the unity? I guess 1 here is the unity and not the number one, but would this part not work without the unity?
But I do not see where they use commutativity?

REPLY [4 votes]: Evaluating $\ a*x = x*a\ $ at $\, x = e\in E\,$ shows that every coefficient $\,a\,$ must commute with every element of the target ring, i.e. the image of $F$ in $E$ is central. It turns out that this necessary condition is sufficient for the evaluation map to be a ring hom.
Remark $\:R\:$-algebras are the rings $\:A\:$ into which one can  evaluate $ R[x]$ for every $\:a \in A\:$. More directly,  an $\:R$-algebra is just a ring $ A\:$ containing a central subring $ R'$ that's a ring image of $\: R\:,\:$ i.e. $\: R'\:$ is either an embedding of $\:R\:$ or $\:R/I\:$ for some ideal $\;I\in R\:.\;$ Being central is precisely the condition needed for elts in $\:R'\:$ to serve as "coefficients" in the sense that this makes polynomial rings $\: R[x]\:$ be  universal $\:R\:$-algebras. Namely, the fact that the coefficients commute with all elts of $\:A\:$ is precisely what is required to make the evaluation map be a ring homomorphism $\: R[x]\to A\:,\:$
viz $\;\; r\; x = x\: r\;$ in $\:R[x]\;\Rightarrow\;  r\: a = a\:r\;\;$ by evaluation $\: x\to a\in A,\:$ i.e. by definition polynomial multiplication assumes that the coefficients commute with the indeterminates, so this property must remain true at values of indeterminates if evaluation is to be a ring homomorphism.