Question about multiple roots of polynomial

Question

Let $R$ be a ring and $f(X) \in R[X], f \ne 0$. Then, $\alpha \in R$ is called multiple root of order $i \in \mathbb{N}^*$ for polynomial $f$ if $(X-\alpha)^i \vert f(X)$ and $(X-\alpha)^{i+1} \not{\vert} f(X)$.

Theorem Let $D$ be an integral domain, $f\in D[X]$.Then, $\alpha \in D$ is a multiple root of order $i$ for $f$ iff there exists $g(X) \in D[X]$ such that $f(X) = (X-\alpha)^i g(X)$ with $g(\alpha) \ne 0$.

I don't understand why there is necesary that $D$ to be integral domain for this to work.

Answer 1

Yeah it is not needed. For the proof we only need that $(X - \alpha)$ is regular, which is clear and holds regardless of the commutative ring. (You should really assume commutativity, otherwise roots behave badly.). Namely, if $(X - \alpha)^{i+1}$ divides $(X - \alpha)^i g$, then $(X - \alpha)$ divides $g$, and the rest follows.

Answer 2

If $\,a\,$ is $\rm\color{#c00}{cancellable},\,$ i.e. $\,ab = ac\iff b = c,\,$ then

$$\begin{align} a^i\,||\,b\!\! \overset{\rm def\!\!}\iff&\ a^i\mid b,\ a^{i+1}\nmid b\\[.3em] \iff &\ b = a^i g,\ \ a^{i+1}\nmid a^i g\\[.3em] \iff &\ b = a^i g,\ \ \color{#0a0}{a\nmid g},\ \text{via $\rm\color{#c00}{cancel}$ $\,a^i$}\\ \end{align}\qquad\qquad$$

OP is the special case $\,a = X\!-\!\alpha \in R[X],\,$ so $\,\color{#0a0}{a\nmid g}\iff X-\alpha\nmid g(X)\iff \color{#0a0}{g(\alpha)\neq 0},\,$ by applying the Polynomial Factor Theorem (here recall that a polynomial is cancellable if it has some cancellable coefficient, e.g. if it is monic).

Beware that the proof fails if $\,a\,$ is not cancellable, and the result may not hold in this case, e.g. $\!\bmod 10\,$ we have $\,2\mid 2\cdot 3,\ 2\nmid 3\,$ but $\,2^2\mid 2\cdot 3\equiv 2^2\cdot 2^2$.