Euclidean Algorithm on $F[X]$

Question

When we do partial fraction decomposition we recall that we can split off coprime factors in the denominator: $$ \frac{a(x)}{p(x)q(x)} = \frac{b(x)}{p(x)} + \frac{c(x)}{q(x)}, \,\,\,\text{where }(p(x), q(x))=1. $$ This because the Euclidean Algorithm allows us to write $1 = c(x)p(x) + d(x)q(x)$. Can someone remind me what allows us to assume that $\deg b < \deg p$ and $\deg c < \deg q$? (Assuming that $\deg a < \deg p + \deg q$.)

Answer 1

First, let's assume that $a(x)/(p(x)q(x))$ is a "proper" fraction (i.e. the degree of $a(x)$ is less than the degree of $p(x)q(x)$). [Without this assumption you can't get the conclusion you're looking for.]

Next, divide $b(x)$ by $p(x)$ and get $b(x)=f(x)p(x)+b_1(x)$ where the degree of $b_1(x)$ is less than that of $p(x)$ (or $b_1(x)=0$). Likewise, $c(x)=g(x)q(x)+c_1(x)$. So we get:

$$\frac{a(x)}{p(x)q(x)} = \frac{b_1(x)}{p(x)}+\frac{c_1(x)}{q(x)}+f(x)+g(x)$$

Clearing denominators: $a(x)=b_1(x)q(x)+c_1(x)p(x)+[f(x)+g(x)]p(x)q(x)$. Therefore, $[f(x)+g(x)]p(x)q(x) = a(x)-b_1(x)q(x)+c_1(x)p(x)$.

Notice that the degrees of $a(x)$, $b_1(x)q(x)$, and $c_1(x)p(x)$ are all strictly less than the degree of $p(x)q(x)$. Therefore, $f(x)+g(x)$ must be zero. :)