Where am I going wrong with this proof for partial fraction decomposition?

Question

We have to prove that $ \frac{F}{G} $ can be written as $\frac{F_1}{G_1}+ \frac{F_2}{G_2}$ if $G=G_1G_2$ and $G_1,G_2$ are co-prime polynomials. This was the proof given in my textbook: $$There \space exists \space polynomials \space C,D \space such \space that \space CG_1+DG_2=1$$ $$Let \space DF=G_1Q+F_1$$ $$Then \space CF=\frac{C}{D}{\left (\frac{Q(1-DG_2)}{C}+F_1 \right)} $$ $$CF= \frac{Q}{D}-QG_2+\frac{C}{D}F_1$$ $$CF+QG_2= \frac{Q}{D}+\frac{C}{D}F_1$$ The next line states that $\frac{Q}{D}+\frac{CF_1}{D}$ can be written as $F_2$. How do we prove that $Q+CF_1$ is divisible by $D$, otherwise we can't make this assumption

Answer 1

That is a very strange proof. It seems easier this way: From $CG_1 + DG_2 = 1$ we have $$ FC\cdot G_1 + FD \cdot G_2 = F $$ Now let $F_1 = FD$ and $F_2 = FC$. Then \begin{align*} \frac{F_1}{G_1} + \frac{F_2}{G_2} &= \frac{F_1 G_2}{G_1 G_2} + \frac{F_2 G_1}{G_2 G_1} \\&= \frac{F_1 G_2 + F_2 G_1}{G_1 G_2} = \frac{F}{G} \end{align*}

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After this answer was accepted I began to think more about the problem, and I decided there must be other conditions not stated. The question title mentions partial fraction decomposition, after all.

I think the full statement is this:

Suppose that $G_1$ and $G_2$ are relatively prime polynomials and $G=G_1G_2$. Suppose further that $F$ is a polynomial with $\deg F < \deg G$. Then there exist polynomials $F_1$ and $F_2$ with $\deg F_1 < \deg G_1$, $\deg F_2 < \deg G_2$, and $$ \frac{F_1}{G_1} + \frac{F_2}{G_2} = \frac{F}{G}$$

I don't like the quoted proof; the use of quotients makes it hard to read. Here is how I would organize it:

Since $G_1$ and $G_2$ are relatively prime, there exist polynomials $C$ and $D$ such that $C G_1 + D G_2 = 1$. By the division algorithm for polynomials, there exist polynomials $Q$ and $F_1$ such that $DF = QG_1 + F_1$, and $\deg F_1 < \deg G_1$. Notice \begin{align*} Q + CF_1 &= (CG_1 + D G_2)Q + CF_1 \\&= CG_1 Q + C F_1 + D Q G_2 \\&= C(QG_1 + F_1) + D Q G_2 \\&= CDF + D Q G_2 = D(CF + QG_2) \end{align*}

Set $F_2 = CF + QG_2$. Notice \begin{align*} F_1 G_2 + F_2 G_1 &= (DF - QG_1)G_2 + (CF + QG_2)G_1 \\&= DG_2 F - QG_1 G_2 + CG_1 F + Q G_1 G_2 \\&= (CG_1 + DG_2)F = F \end{align*} Therefore the equation $ \frac{F_1}{G_1} + \frac{F_2}{G_2} = \frac{F}{G}$ is satisfied.

It remains to show that $\deg F_2 < \deg G_2$. Since $F_2 G_1 = F - F_1 G_2$, we know that $$ \deg(F_2 G_1) \leq \max \{\deg F, \deg(F_1 G_2)\} $$ Now \begin{align*} \deg(F_1 G_2) &= \deg F_1 + \deg G_2 < \deg G_1 + \deg G_2 \\ \deg F &< \deg G = \deg G_1 + \deg G_2 \end{align*} So in either case, $$ \deg F_2 + \deg G_1 < \deg G_1 + \deg G_2 $$ and we can cancel $\deg G_1$ from both sides.

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Here is an example which compares the approaches. You can see that the second is more “interesting” than the first.

Let $G_1 = x+1$, $G_2 = x^2 + 1$, and $F = x$. So we're looking for a decomposition of $\frac{x}{(x+1)(x^2+1)}$. The Bézout coefficients $C = \frac{1}{2}(1-x)$ and $D = \frac{1}{2}$ satisfy $CG_1 + DG_2 = 1$.

In the first answer, we let $F_1 = FD = \frac{1}{2} x$ and $F_2 = FC = \frac{1}{2}x(1-x)$. The resulting decomposition is $$ \frac{x}{(x+1)(x^2+1)} = \frac{\frac{1}{2} x}{x+1}+ \frac{\frac{1}{2}x(1-x)}{x^2+1} $$ Both of the fractions on the right are “improper” in the sense that the degree of the numerator equals that of the denominator.

In the second answer, we divide: $$ DF = \frac{1}{2} x = \frac{1}{2}(x+1) - \frac{1}{2} $$ So $Q=\frac{1}{2}$ and $F_1 = - \frac{1}{2}$. Then $$ F_2 = CF + QG_2 = \frac{1}{2}(1-x)x + \frac{1}{2}(x^2+1) = \frac{1}{2} x + \frac{1}{2} $$ Check: \begin{align*} F_1 G_2 + F_2 G_1 = -\frac{1}{2}(x^2+1) + \frac{1}{2}(x+1)(x+1) = x \end{align*} This time the decomposition is $$ \frac{x}{(x+1)(x^2+1)} = \frac{-\frac{1}{2}}{x+1}+ \frac{\frac{1}{2}(x+1)}{x^2+1} $$ and both fractions on the right are “proper.”

Answer 2

Say, we are working in some polynomial ring $K[x]$ over a field $K$.

We know that the left-hand side of the equation is the sum of the product of respectively two polynomials in $K[x]$. Hence the left-hand side as a whole is a polynomial in $K[x]$, which implies that $Q+CF_1$ must be divisible by $D$ for otherwise the right-hand side would not be a polynomial in $K[x]$.