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Theorem 6.6 in Hungerford's Algebra book: Let $R$ be a commutative ring with identity and $f \in R[x]$. Then $c \in R$ is a root of $f$ is and only if $x-c$ divides $f$.

Here is the proof he offers:

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

Here is corollary 5.6:

If $\varphi : R \to S$ is a homomorphism of commutative rings and $s_1,s_2,...,s_n \in S$, then the map $R[x_1,...,x_n] \to S$ given by $f \mapsto \varphi f(s_1,...,s_n)$ is a homomorphism of rings.

This seems more complicated than it ought to be. If $x-c \mid f(x)$, then $f(x) =h(x)(x-c)$. Evaluating $f$ at $c$ (i.e., applying the homomorphism $x \mapsto c$), we get $f(c)=h(c)(c-c)=0$, and we are done. What's wrong with doing that?

user193319
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  • "Evaluating" works because of Corollary 5.6. Division with remainder is necessary for the opposite direction. – darij grinberg Dec 31 '17 at 17:56
  • @darijgrinberg I understand that. But why apply it to $(h(x)-q(x))(x-c)=f(c)$, which requires algebra/manipulation to get that equation (albeit simple algebra/manipulation) rather than simply apply the homomorphism to $f(x) = h(x)(x-c)$? – user193319 Dec 31 '17 at 17:59
  • Oh. Yes, he makes a detour. Perhaps it is useful for later reference (not sure). – darij grinberg Dec 31 '17 at 18:08

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As Corollary 5.6 from your book states, the problem is in the commutativity requirement. A simple example where substitution fails is to look at $p(x)=x^2+1$ over the Quaternions $\mathbb{H}$. It is easy to see that $i,j,k$ are roots of this polynomial, i.e.

$$p(i)=p(j)=p(k)=0$$

But if you factor out

$$p(x)=(x+i)(x-i)$$

and substitute $j$ for instance

$$p(j)\overset{?}{=}(j+i)(j-i)=j^2+ij-ji-i^2=2ij\neq 0$$

you certainly don't get zero. Thus in fact

$$p(j)\neq(j+i)(j-i)$$

You can look at it this way: when you write $p(x)=(x+i)(x-i)$ you assume $x$ is in the centralizer of $\mathbb{H}$, but when you plug in $x=j$ that's false.

This answer is based on an answer to another stack-exchange post. The credit goes to @ArturoMagidin.

eranreches
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