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We know that :

If $K$ is a field and $P\in K[X]$ is a polynomial with coefficients in $K,$ then $\alpha\in K$ is a root of $P$ if and only if $(X-\alpha)$ divides $P$.

My question is the following : does this hold for polynomials with coefficients in an integral domain $R,$ and if it doesn't can you give me a counterexample ? I tried to think about the field of fractions, but no managed to find an answer in the general case. Thank you for all your answers !

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  • $\begingroup$ Yes of course.. $\endgroup$
    – MSE
    Jan 15, 2016 at 22:07
  • $\begingroup$ Yes; in fact it holds when the coefficient ring is any commutative ring with 1. The way to show it is use a generalized polynomial division algorithm. $\endgroup$
    – ebrahim
    Jan 15, 2016 at 22:09
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    $\begingroup$ @ebrahim : thank you for your answer, I did not know that we can do that kind of division - which works here because the coefficient in front of the $X$ is one and so invertible if I well understood ! $\endgroup$
    – Balloon
    Jan 15, 2016 at 22:13

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As already indicated in the comments the answer is yes. The easiest way to see this is to note that $$x^k-\alpha^k=(x-\alpha)(x^{k-1}+\alpha x^{k-2}+\alpha^2x^{k-3}+\cdots+\alpha^{k-1})$$ so for $f(x)=c_nx^n+\cdots+c_1x+c_0$ it is clear that $$f(x)-f(\alpha)=c_n(x^n-\alpha^n)+\cdots+c_1(x-\alpha)$$ is divisible by $x-\alpha$. Therefore, $f(x)$ is divisible by $x-\alpha$ precisely when $f(\alpha)=0$.

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  • $\begingroup$ Nice way to prove it, +1 and thank you ! $\endgroup$
    – Balloon
    Jan 15, 2016 at 22:19
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Yes it does, let $D$ be an integral domain with a unit.

Then it can be shown via induction (over the degree of the polynomial) that generalized division holds.

In other words, given $P$ a polynomial and $D$ a monic polynomial, polynomials $Q$ and $R$ exist so that:

$P=QD+R$, where the degree of $R$ is less than the degree of $D$.


Going back to our problem, let $P$ be the polynomial, and let $\alpha$ be the proposed root.

We let $D=(x-\alpha)$ and apply generalized division to get:

$P=(x-\alpha)Q+c$, where $c$ is in the integral domain (because the degree is $1$).

Clearly when we evaluate the polynomial at $x$ we get:

$P(x)=0+c$. So $\alpha$ is a root of $P$ if and only if $c=0$, if and only if $(x-\alpha)$ divides $P$.

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  • $\begingroup$ Does that not only work when the polynomial ring is a euclidean domain? namely when F is a field then F[x] is an ED? $\endgroup$
    – snulty
    Jan 15, 2016 at 22:30
  • $\begingroup$ No, you don't need the ring of polynomials to be a euclidean domain, it suffices to have division by monic polynomials for the result. $\endgroup$
    – Asinomás
    Jan 15, 2016 at 22:37
  • $\begingroup$ But it need to be an ED to apply the division algorithm no? $\endgroup$
    – snulty
    Jan 15, 2016 at 22:39
  • $\begingroup$ Or is that just special for monic polynomials? $\endgroup$
    – snulty
    Jan 15, 2016 at 22:40
  • $\begingroup$ It is special for monic polynomials, the key thing is that, the leading coefficient is $1$, so we don't need to do division. When you try to divide by a polynomial that is not monic you may run into the necesity of inverses, but in this case it's all good. $\endgroup$
    – Asinomás
    Jan 15, 2016 at 22:41

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