Recently, I find some question about this answerFinitely generated graded modules over $K[x]$, although I want to comment on original post, the author say that he will not post anything anyway.
The first part is:
$1.$"since it is a graded PID they are of the form $(x^n)$" Does he means that every ideal in graded PID ring is always this form? I don't think so, since we only need principal ideal.
$2.$ In surjective homomorphism:$\varphi$ $$\varphi: K[x]y_1 \oplus \cdots\oplus K[x]y_n\rightarrow \Sigma^{\deg(y_1)}K[x]/(x^{q_1}) \oplus \cdots\oplus \Sigma^{\deg(y_m)}K[x]/(x^{q_m}) \bigoplus_{i=1}^{n-m} \Sigma^{\deg(y_{m+i})}K[x]$$
by mapping $(\alpha_1y_1,...,\alpha_my_m) \rightarrow (\alpha_1\bmod(x^{q_1}),...,\alpha_m\bmod(x^{q_m}),\alpha_{m+1},...,\alpha_{n})$
Does $n$-shift upward in grading means transfer $H=\bigoplus_{i\in \Bbb Z}H_i$ to $H'=\bigoplus_{i-deg (y_j)}H_i$ , also what is the changes? I mean, just shift, $\Sigma^{\deg(y_1)}K[x]/(x^{q_1})$ is isomorphic (as module) to $K[x]/(x^{q_1})$ why we need shifted?
$3.$ Does the $deg(y_j)$ means homogeneous degree(ie in the graded ring $(K[x])^n$, we have graded $\oplus_{i\in \Bbb Z_{\ge 0}} \oplus_{l\in{1...n}}(x^i K))$(negative part is all zero), so $deg(y_j)$ means $y_j \in \oplus_{l\in{1...n}} x^{deg(y_j)}K$, but how can we show this ?
I know that My problems may be too much, but if you have some answer or idea, please tell me, because I really want to understand what is going on. I will very appreciate anyone who give me advise.
