Let $\Bbb F$ be a field and $M$ a finitely generated $\Bbb F[x]$-module. The structure theorem for modules over a PID says that $$ M\cong \Bbb F[x]^r\oplus\biggl(\bigoplus_{j=0}^s\Bbb F[x]/(f_j(x))\biggr)\text, $$ where $r$ and $s$ are non-negative integers and $f_j(x)$ are polynomials in $\Bbb F[x]$.
Now, because the ring $\Bbb F[x]$ is graded by the standard grading $\Bbb F[x]=\sum_{i=0}^\infty x^i\Bbb F$, according to this paper (Theorem 4.8.), $M$ can be written also in the form $$ M\cong\biggl(\bigoplus_{i=1}^M(x^{a_i})\biggr)\oplus\biggl(\bigoplus_{j=1}^N(x^{b_j})/(x^{c_j})\biggr) $$ for some non-negative integers $M,N$, $a_i,b_j,c_j$.
I can't understand this sentence from the proof:
The free component is composed of graded rings of the form $\bigoplus_{i\geq q}x^i\Bbb F$ , which are isomorphic to ideals of the form $(x^q)$.
The free part is $\Bbb F[x]^r$: How can I write this as a direct sum of things of the form $\bigoplus_{i\geq q}x^i\Bbb F$?
The free part is a direct sum of shifted copies of $\mathbb F[X]$. That is, $\mathbb F[X](-k)$ for some $k\ge 0$. This is nothing but $X^k\mathbb F[X]$.