# Noncommutative rings and the evaluation homomorphism

Recall the evaluation homomorphism of a ring. For example, if $\mathbb R[x]$ is the ring of polynomials with real coefficients then we can evaluate a polynomial $p(x)=a_0+a_1x+\cdots+a_nx^n$ with respect to $c$ by letting $p(c)=a_0+a_1c+\cdots+a_nc^n$. Note that this operation is a ring homomorphism. Now it is a theorem that this is not a homomorphism in general for noncommutative rings. Anyone can give an explicit counterexample? Thanks.

• What example did you try? It is quite difficult to stumble upon examples in which the function is a ring morphism, in fact! I am quite sure that, with a bit of luck, if you pick pretty much any noncommutative ring R you'd find an example yourself... Feb 9, 2016 at 5:34
• You are missing something in what you wrote. Otherwise you say that «note that this operation is a ring homomorphism» followed immediately by «it is a theorem that this is not a homomorphism in general», and that makes no sense. Feb 9, 2016 at 5:36
• I meant something like: "However, in the noncommutative case, it is a theorem that this is not a homomorphism in general"
– user310648
Feb 9, 2016 at 6:04

If evaluation $\text{ev}_r : R[x] \to R$ were a homomorphism, then it would have to satisfy
$$\text{ev}_r(x \cdot a) = \text{ev}_r(ax) = ar = \text{ev}_r(x) \text{ev}_r(a) = ra$$
for every $a \in R$: that is, $r$ would have to be central. And in fact $x$ is central in $R[x]$, so there is an evaluation homomorphism $\text{ev}_r : R[x] \to R$ precisely when $r$ is central.