Math and science::Algebra::Aluffi

Rings. Uniqueness of polynomial division.

Two versions of the same idea: unique polynomial division.

Let $R$ be a ring, $R [x]$ be a polynomial ring and let $f (x), g (x) \in R [x]$ be polynomials, with $f (x)$ being a monic polynomial.

Then there exists a unique pair of polynomials $q (x), r (x) \in R [x]$ where $\deg r < \deg f$ such that:

g (x) = q (x) f (x) + r (x)

In terms of cosets:

Let $R$ be a ring, $R [x]$ be a polynomial ring and let $f (x), g (x) \in R [x]$ be polynomials, with $f (x)$ being a monic polynomial.

Then there exists a unique polynomial $r (x) \in R [x]$ where $\deg r < \deg f$ such that:

g (x) + (f (x)) = r (x) + (f (x))

These theorems rely on a lemma showing polynomial long division to be unique when $f (x)$ is monic. Can you remember the proof?

The generator notation $(f (x))$ is used to denote the ideal generated by $f (x)$ .

Long division is unique. Proof.

Let $R$ be a ring, $R [x]$ be a polynomial ring and let $f (x), g (x) \in R [x]$ be polynomials, with $f (x)$ being a monic polynomial. Suppose that there exist polynomials $q_{1} (x), q_{2} (x), r_{1} (x), r_{2} (x) \in R [x]$ such that $\deg r_{1} (x)$ and $\deg r_{2} (x)$ are less than $\deg f (x)$ and that:

g (x) = q_{1} (x) f (x) + r_{1} (x) = q_{2} (x) f (x) + r_{2} (x)

Our goal is to show that $q_{1}$ must be equal to $q_{2}$ and that $r_{1}$ must be equal to $r_{2}$ .

Rearranging, we have:

(q_{1} (x) - q_{2} (x)) f (x) = r_{2} (x) - r_{1} (x)

Proof by contradiction: assume that $r_{1}! = r_{2}$ . $f (x)$ being monic means that the LHS has degree greater than $f (x)$ , which is a contradiction, as the RHS is asserted to have degree less than $f (x)$ . So $r_{1} (x) = r_{2} (x)$ and $q_{1} (x) = q_{2} (x)$ .

Note how the above proof required $f (x)$ to be monic.

Why monic? Continued.

Having the multiplicative identity in the leading coefficient ensures that the leading coefficient is not a zero divisor. In fact, it seems that any unit should be fine. Having a non-zero divisor ensures that the remainder has degree strictly less than the degree of $f (x)$ . If $g (x) = x$ and $f (x) = 2 x$ , then $q (x) = 0$ and $r (x) = x$ is a valid solution, but $r (x)$ has degree equal to the degree of $f (x)$ . This is a problem because we will rely on recusion with decreasing degree to prove subsequent results.

Source

Aluffi p147, p148