# Partial fraction decomposition over arbitrary field.

Let $K$ be a field and $f,g$ be (1 variable) polynomials over $K$, and suppose that $g=p_{1}^{e_1} p_{2}^{e_2} \cdots p_{k}^{e_k}$ where each $p_i$ is irreducible over $K$ and $e_i \geq 1$. Does there exist polynomials $b$ and $a_{ij}$ with the following properties?

• $\deg{a_{ij}}<\deg{p_i}$ for all $i=1,\ldots,k$
• $\displaystyle \frac{f}{g}=b+\sum_{i=1}^{k}\sum_{j=1}^{e_k}\frac{a_{ij}}{p_{i}^{j}}$

Moreover, are such polynomials unique?

What I have tried: Since $\{p_{i}^{e_i}\}$ are pairwise relatively prime, there are polynomials $A_1,\ldots,A_k$ (Bezout) such that $$A_1 p_1^{e_1}+\cdots+ A_k p_k^{e_k}=1$$ and thus we may write $\displaystyle \frac{f}{g}=\frac{fA_1 p_1^{e_1}+\cdots+ fA_k p_k^{e_k}}{g}=\frac{fA_1}{p_2^{e_2}\cdots p_k^{e_k}}+\cdots+\frac{fA_k}{p_1^{e_1}\cdots p_{k-1}^{e_{k-1}}}$. Repeating this on every summand $k$ times, we get polynomials $B_i$ such that $\displaystyle \frac{f}{g}=\sum_{i=1}^{k}\frac{B_i}{p_{i}^{e_i}}$, and after long division (if necessary) there exist polynomials $b,\tilde{B}_i$ such that $$\frac{f}{g}=b+\sum_{i=1}^{k}\frac{\tilde{B}_i}{p_{i}^{e_i}}$$ with $\deg{\tilde{B}_i}<\deg{p_i^{e_i}}$.

Where I'm stuck: I don't see how I should proceed with the summands of the form $\frac{\tilde{B}_i}{p_i^{e_i}}$. Since $\{p_i,p_i^2,\ldots,p_i^{e_i}\}$ are not relatively prime, Bezout does not work and I don't see how to choose $a_{ij}$ from $\tilde{B}_i$ (unless $p_i^{e_i}$ is linear...). I'm also having difficulties with the uniqueness part of the assertion.

Can someone give me an advice for this problem? Please enlighten me!

You have reduced the problem to the case where there is only one irreducible factor. Let's write it as $\frac{f}{g^n}$ where $g$ is irreducible in $K$, and $\deg f < \deg g^n$. We seek polynomials $a_1, a_2, \ldots, a_n$, with $\deg a_i < \deg g$, such that $$\frac{f}{g^n} = \sum_{i=1}^{n} \frac{a_i}{g^i}.$$
If we clear denominators, we get this (where I have purposely split out the last term because it has no $g$'s): $$f = \left(\sum_{i=1}^{n-1} a_i g^{n-i}\right) + a_n.$$
Now the $a_i$'s can be computed by successive division. The polynomial $a_n$ is the remainder when $f$ is divided by $g$, then $a_{n-1}$ is the remainder of $(f-a_n)/g$, etc.
• Can't we just let $a_n$ be the remainder of $f$ when divided by $g$? and let $a_{n-1}$ be the remainder of $(f-a_n)/g$. . . so on? – Dilemian Jan 13 '17 at 19:46
• @Dilemian Yes, you are correct. My original answer was incorrect because I thought $\deg a_i$ had to be less than $i$ for some reason. It should be $\deg a_i < \deg g$. – Ted Jan 14 '17 at 6:29