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.


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 $x - a \mid x^k - a^k$.

  • 1
    $\begingroup$ Lovely - thanks. $\endgroup$ – John Gowers Feb 5 '14 at 17:18

You can do polynomial long division over any commutative ring, as long as the leading coefficient of the divisor is a unit. Think about the process - at each step, you only need to divide the leading coefficient of the divisor into the leading coefficient of the polynomial you have left (after multiplying and subtracting from the previous step).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.