I have to prove that for $f,g\in F[X]$, $F$ a field, where $g$ is monic with prime factorization $g=g_1^{v_1}...g_n^{v_n}$ (where the $g_i$'s aren't pairwise associated prime elements), there is an unique way in $F(X)$ to write $f/g = f_o + \sum\limits_{i=1}^n\sum\limits_{j=1}^{v_i}c_{ij}/g_i^j$ with polynomials $f_o,c_{ij}\in F[X]$ where $\deg(c_{ij})<\deg(g_i)$.
Now $f_o$ I can get through polynom divison: $f=f_og+r$.
Now first I want to show that I can write $f/g=f_0+\sum\limits_{i=1}^nf_i/g_i^{v_i}$, with $\deg f_i<\deg g_i^{v_i}$. I try to use induction over $n$. For the case $n=1$ it is clear with $f_1=r$. So let $n>1$. I can write $g=\bar gg_1^{v_1}$ with $\bar g =g_2^{v_2}...g_n^{v_n}$. Now I need to show that $r=h\bar g + kg_1^{v_1}$ with unique $h,k\in F[X]$ with appropriate degrees ($\deg(h)<\deg(g_1^{v_1})$ and $\deg(k)<\deg(\bar g)$). Im not sure how to preceed, or if this is the right way!
For the last bit I have to show that $f_i=\sum a_jg_i^j$ uniquely with $\deg a_j<\deg g_i$. The existence here is easy: I just use polynom division $f_i=q_1g_i^{\lfloor \deg f_i/\deg g_i\rfloor} +r_1$, then $r_i=q_{i+1}g_i^{\lfloor \deg r_{i}/\deg g_i\rfloor} +r_{i+1}$. But how do I show uniqueness? edit: Ok, I got uniqueness just now! (Because if $f_i=\sum a_jg_i^j=\sum b_jg_i^j$ let $j_0$ be highest s.t. $a_{j_0}\neq b_{j_0}$ then $\sum\limits_{n\geq j_o} (a_j-b_j)g_i^j=cX^d$ for some $c\in F$ and $n\in \mathbb N:n\geq j_0\deg(g_i)$, which can't dissapear with the lower terms of the sum as they are of degree less than $j_0\deg(g_i)$.
