# Proof of the single factor theorem over an arbitrary commutative ring

Theorem (Single factor theorem) Let $R$ be a commutative ring, and let $P\in R[X]$, where $R[X]$ is the polynomial ring over the indeterminate $X$. Suppose $P(\alpha)=0$. Then $X-\alpha$ divides $P(X)$.

If $R$ is a field, then I know the following proof:

Proof (for a field): Using polynomial long division, we can write:

$$P(X)=Q(X)(X-\alpha)+r$$

where $r\in R$ is a constant, since it has to have strictly lower degree than $X-\alpha$. But then, putting $X=\alpha$, we get $0=0+r$, so $r=0$, and we win. $\Box$

The proof above doesn't work over an arbitrary ring, because we cannot always do polynomial long division over an arbitrary ring. However, I have read in a number of places (including on this site) that the theorem is true when $R$ is an arbitrary commutative ring. I haven't been able to find anything other than the proof for fields outlined above, either online or in Chrystal's Algebra. One idea I had was to try and use Gauss's lemma, but I'd like to know if there is a 'standard' proof of this fact.

• Dividing by $x-\alpha$ also works in an arbitary ring because the leading coefficient of $x-\alpha$ is 1. – Peter Feb 5 '14 at 17:19
• – Martin Sleziak Jun 26 at 17:58

Let $p(x) = c_n x^n + \cdots + c_1 x + c_0$. Then: $$p(x) = p(x) - p(a) = c_n (x^n - a^n) + \cdots + c_1 (x - a)$$ But over any ring $x - a \mid x^k - a^k$.