Here is the question I am reading the answer of @Xam to it, but I am wondering why
Prove that if $R$ is an integral domain and has ACCP, then $R[X]$ has ACCP
1-I am wondering in his answer in the second paragraph, specifically when he said "As $P_{n+i+1}\mid P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R$." why he said for some $r_{i} \in R$ and not for some $r_{i} \in R[X],$are not we speaking about divisibility of 2 polynomials? could anyone explain that to me please?
2-Also, I did not get the relation between the two leading coefficients in the paragraph following it. why they should be related? the two polynomials could have the same degree but the leading coefficients no one of them is a multiple of the other. could anyone explains this also to me?
3-My last question, why we are adding $n$ to $k,$ why we need to do that? can not $k$ be inside $n$?
He had concluded before that $\deg P_n = \deg P_n+i$ for all $i\in \mathbb{N}$. Now as $P_{n+i+1}∣P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R[X]$. But then $$\deg P_{n+i} = \deg P_{n+i+1}=\deg r_i + \deg P_{n+i+1}$$ (here we used that $R$ is an integral domain). This leads to the conclusion that $r_i$ is constant and hence $r_i\in R$.
As for the relation of the leading coefficients: If $P_{n+i}=r_iP_{n+i+1}$ and the corresponding leading coefficients are $a_{n+i}$ (for $P_{n+i}$) and $a_{n+i+1}$ (for $P_{n+i+1}$), then, by the definition of polynomial multiplication, we must have $a_{n+i}= r_ia_{n+i+1}$.
At $n$, the degree becomes stationary allowing for the argument above. Then, maybe at a later time, the chain on $R$ becomes stationary. The rest of the argument needs that both are stationary from some index on. The chain of ideals generated by leading coefficients could become stationary earlier while stile having a varying degree (think of multiplication with monic polynomials).