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Definition: A nonzero element $a$ of a commutative ring $R$ is said to divide an element $b\in R$ (notation: $a\mid b$) if there exists $x\in R$ such that $ax= b$.

Let $R$ be a commutative ring with identity and $f\in R[x]$. Then $c\in R$ is a root of $f$ if and only if $x - c$ divides $f$.

Sketch of proof: We have $f(x) = q(x)(x - c)+f(c)$ by Corollary 6.3. If $x - c\mid f(x)$, then $h(x)(x - c) = f(x) = q(x)(x - c) +f(c)$ with $h\in R[x]$, whence $(h(x) - q(x))(x - c) = f(c)$. Since $R$ is commutative, Corollary 5.6 (with $\varphi = 1_R$) implies $f(c) = (h(c) - q(c))(c - c) = 0$. Commutativity is not required for the converse; use Corollary 6.3.

Author claims that commutativity is not needed to show: $c\in R$ is root of $f$$\implies$$x-c$ divides $f$. If we don’t assume $R$ to be commutative, then $R[x]$ may not be commutative as well. And the notion of divides don’t make sense. Am I missing some subtlety?

Author given proof of, $x-c$ divides $f$ $\implies$ $c\in R$ is root of $f$, is not optimised (see here). I have not written corollary 6.3 because it is in the first sentence of proof.

user264745
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    Good point. Also the requirement in the definition that $a$ is non-zero is not reasonable. It implies a cascade of unnecessary case distinctions (for example, even here, $x-c$ can be the zero element, namely when $R=0$). See also https://mathoverflow.net/questions/45951/interesting-examples-of-vacuous-void-entities – Martin Brandenburg Jan 01 '24 at 16:52
  • @MartinBrandenburg Thank you for the link. – user264745 Jan 01 '24 at 18:09
  • Isn't your question a duplicate of the already mentioned https://math.stackexchange.com/questions/2586854 ? – Martin Brandenburg Jan 01 '24 at 18:36
  • @MartinBrandenburg I don’t think so. That post concern with, $x-c$ divides $f$ $\implies$ $c\in R$ is root of $f$, part of the proof (slightly better version of author proof). – user264745 Jan 01 '24 at 18:39
  • The answer there shows that the other implication is wrong in the non-commutative case, thus answering your question and disproving Hungerford's statement. – Martin Brandenburg Jan 01 '24 at 18:45
  • @MartinBrandenburg To be honest, I didn’t understand answer because quaternions ring is used. So I pre assumed, that answer show statement “if $f=gh$ ($f,g,h\in R[x]$) and $c\in R$, then $f(c)=g(c)h(c)$” is false if $R$ is not commutative. – user264745 Jan 01 '24 at 18:50

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