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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 \( \operatorname{deg} r < \operatorname{deg} f \) 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 \( \operatorname{deg} r < \operatorname{deg} f \) 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 \( \operatorname{deg} r_1(x) \) and \( \operatorname{deg} r_2(x) \) are less than \( \operatorname{deg} f(x) \) 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 \( f(x) = 2x\), 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