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 $X^k\mathbb F[X]$.

  • $\begingroup$ What do you mean exactly by $\Bbb F[x](-k)$? I can't understand how the sum could be direct if the summands overlap $\endgroup$ – User Feb 2 '16 at 16:56
  • $\begingroup$ Notation is explained e.g. here. $\endgroup$ – user26857 Feb 2 '16 at 16:57
  • $\begingroup$ Btw, you got a long explanation of the shifting phenomenon in a former question: math.stackexchange.com/questions/1610978/… $\endgroup$ – user26857 Feb 2 '16 at 17:00
  • $\begingroup$ Now I understand the notation $\Bbb F[x](-k)$ which is the same as $\Sigma^k\Bbb F[x]$ of my other question, but I still can't understand the direct sum. If the free part has rank 1, then it is $\Bbb F[x]$ which is the same as $\Bbb F[x](-0)$. But if, for example, the free part has rank 2, $\Bbb F[x]\oplus \Bbb F[x]$, how can I write it in terms of $\Bbb F[x](-k)$'s? What are the values of $k$? $\endgroup$ – User Feb 2 '16 at 17:22
  • $\begingroup$ The shifts appear as follows: if $M$ is free of rank one and the homogeneous generator is $z$ with $\deg z=d>0$, then $M\simeq \mathbb F[X](-d)$. Why? Because the (iso)morphisms of graded modules are usually graded, that is, send $n$th degree part into $n$th degree part. In the example above, $1\mapsto e$ is not a graded morphism since $\deg 1=0$ and $\deg z=d$. Then you have to make $\deg 1=d$ by moving it $d$ places to the right. This means to consider $\mathbb F[X](-d)$ which has $\mathbb F[X](-d)_n=\mathbb F[X]_{n-d}$, that is, $\endgroup$ – user26857 Feb 2 '16 at 17:31

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