I found that my 2 textbooks and other answers state this theorem for commutative ring:
- From textbook Algebra by Saunders MacLane and Garrett Birkhoff.
- From textbook Analysis 1 by Herbert Amann and Joachim Escher.
- In this answer, @Bill Dubuque also states for commutative ring.
For polynomials over any commutative coefficient ring, the high-school polynomial long division algorithm works to divide with remainder by any monic polynomial…
- In this answer, @Bill Dubuque also states for commutative ring.
Yes, your intuition correct: the Polynomial Factor Theorem works over any commutative ring since we can always divide (with remainder) by a polynomial that is monic i.e. lead coef $=1$ (or any unit = invertible element). Ditto for the equivalent Polynomial Remainder Theorem – see below.
I have re-read the proofs in my 2 textbooks and can not found where the commutativity is used. As such, is commutativity needed in the proof of division algorithm?
Best Answer
If $K$ is noncommutative there's a bit of ambiguity in what you mean by $K[x]$; if you mean that $x$ is central, then I suppose the argument goes through fine, but I'm not aware of any applications of it.
Note that it does not follow that polynomials over $K[x]$ have unique factorization. As an explicit example, if $K = \mathbb{H}$ is the quaternions, then the polynomial $x^2 + 1$ admits inequivalent factorizations
$$x^2 + 1 = (x + i)(x - i) = (x + j)(x - j) = (x + k)(x - k).$$
The problem is that because we need $x$ to be central, we can't substitute any non-central element for $x$! So the usual argument where we substitute a root no longer applies.