# Question on uniqueness in the proof used for the partial fraction algorithm

Partial fraction algorithm

Let $\mathbb{F}$ be a field, and let $f(x)$, $a(x)$, $b(x)$ be polynomials in $\mathbb{F}[x]$ such that $a(x)$ and $b(x)$ are coprime and $\deg f < \deg a + \deg b$. Then there exist unique polyniomials $r(x)$ and $s(x)$ with $\deg r < \deg a$, $\deg s < \deg b$ such that $$\frac{f(x)}{a(x) b(x)} = \frac{r(x)}{a(x)} + \frac{s(x)}{b(x)}.$$

I think existence of the polynomials can be shown as follows:

As $a(x)$ and $b(x)$ are coprime they will have no factors in common. Thus $\gcd (a(x),b(x)) = 1$. Now from Bezout's identity there exist polynomials $g(x)$ and $h(x)$ such that $$1 = a(x) \cdot g(x) + b(x) \cdot h(x).$$

Now \begin{align*} \frac{f(x)}{a(x) b(x)} &= \frac{f(x) \cdot 1}{a(x) b(x)}\\ &= \frac{f(x) \left (a(x) \cdot g(x) + f(x) \cdot b(x) \right )}{a(x) b(x)}\\ &= \frac{f(x) \cdot a(x) \cdot g(x)}{a(x) b(x)} + \frac{f(x) \cdot b(x) \cdot h(x)}{a(x) b(x)}\\ &= \frac{f(x) \cdot h(x)}{a(x)} + \frac{f(x) \cdot g(x)}{b(x)} \end{align*} And if we let $r(x) = f(x) \cdot h (x)$ and $s(x) = f(x) \cdot g(x)$ this gives $$\frac{f(x)}{a(x)b(x)} = \frac{r(x)}{a(x)} + \frac{s(x)}{b(x)},$$ and shows existence.

Questions

1. How do I know $\deg r < \deg a$ and $\deg s < \deg b$?

2. How does one go about showing the uniqueness of $r(x)$ and $s(x)$?

Your original $r(x)$ might not have small enough degree; but the division algorithm allows you to write $r(x) = t(x) a(x) + u(x)$ where $\deg u < \deg a$. Writing $v(x) = s(x) - t(x) b(x)$ gives $$\frac{f(x)}{a(x)b(x)} = \frac{u(x)}{a(x)} + \frac{v(x)}{b(x)};$$ can you verify that $\deg v < \deg b$?
As for uniqueness, try subtracting two such representations from each other; you'll get $$\frac{r(x)-r'(x)}{a(x)} = \frac{s'(x)-s(x)}{b(x)}.$$ By coprimality of $a(x)$ and $b(x)$, you should be able to prove that $a(x)$ divides $r(x)-r'(x)$ (and similarly for the other fraction); and this will imply the required uniqueness.