Why we need to differentiate between the deg. of the polynomial being stationary and the leading coefficient being stationary?

Question

Here is the question I was reading the answer to it by Xam:

Prove that if $R$ is an integral domain and has ACCP, then $R[X]$ has ACCP

Here is the answer by Xam:

This answer is for item ii) given in the question. The other answer is fine, with the corrections given in the comments, but there is not necessary at all to use the fact that $F[X]$ satisfies the ACCP.

As usual, let $$(P_1)\subseteq (P_2)\subseteq \ldots$$ be a chain of principal ideals of $R[X]$. Then $P_{i+1}\mid P_i$ for all $i\ge 1$, and this implies that $\deg(P_1)\ge \deg(P_2)\ge \ldots$ is a decreasing sequence of natural numbers, then it must stabilize and thus there is some $n\in \Bbb{N}$ such that $\deg(P_n)=\deg(P_{n+i})$ for all $i\ge 0$. As $P_{n+i+1}\mid P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R$.

Now, let's denote $a_i$ the leading coefficient of $P_{n+i}$, therefore $a_i=r_ia_{i+1}$ for all $i\ge 0$ and then $a_{i+1}\mid a_{i}$, which lead us to the following chain of principal ideals of $R$: $$(a_0)\subseteq (a_1)\subseteq \ldots$$ Since $R$ satisfies the ACCP, then there exists $k\in \Bbb{N}$ such that $(a_k)=(a_{k+j})$ for all $j\ge 0$. Let's set $t=n+k$, we claim that $(P_t)=(P_{t+j})$ for all $j\ge 0$. Indeed, since $P_{t+j}\mid P_t$, it's enough to prove that $P_t\mid P_{t+j}$. Let's write $P_t=dP_{t+j}$, with $d\in R$, then $da_{k+j}=a_k=ua_{k+j}$, with $u\in R^{\times}$ because $a_k\sim a_{k+j}$. As $a_{k+j}\neq 0$, we deduce that $d=u$. Therefore $d^{-1}P_t=(d^{-1}d)P_{t+j}=P_{t+j}$, so $P_t\mid P_{t+j}$. Hence, $R[X]$ satisfies the ACCP.

I am asking about the last paragraph:

Why we need to differentiate between the deg. of the polynomial being stationary and the leading coefficient being stationary? could anyone explain that to me please?

NOTE: Below is part of a hint from @Bill Dubuque in my previous question about this step but still I am not understanding:

Finally by R a domain: if $0≠P,Q∈R[x]$ have equal degree and $P∣Q$ then they are associate ⟺ their lead-coefs are associate (and divisibility minimal elements are associate since they must divide each other).

Answer 1

You do not need stationary degrees to conclude abot the leading coefficients. Ignoring anything about degrees, let $a_k$ be the leading coefficient of $P_k$. As $P_k\in(P_{k+1})$, we know that $P_k=QP_{k+1}$ for some $Q$ (with $\deg Q=\deg P_k-\deg P_{k+1}$) and by looking at the leading coefficients we see that $a_k\in(a_{k+1})$. So we have our ascending chain $$ (a_1)\subseteq (a_2)\subseteq (a_3)\subseteq \ldots$$ even without considering degrees. As this gets stationary, we find $n$ such that $(a_k)=(a_n)$ for all $k\ge n$. In particular $a_k=r_ka_n$ for some $r_k\in R^\times$. Now it follows for example for all $k\ge n$ that $X^{\deg P_n-\deg P_k}P_k-r_kP_n$ is $\in (P_k)$ and due to cancellation of leading terms is of degree $<\deg P_n$.

However, this does not help us at all because having degree $<\deg P_n$ is nothing unusual when all we can say about the general non-zero element of $(P_k)$ is that it has degree $\ge \deg P_k$; after all, there is nothing that prevents $\deg P_k$ from being smaller than $\deg P_n$. Or is there?

Cue the non-increasing and hence eventually stationary degrees. That is, by perhaps switching to some larger $n$, we can ensure that $\deg P_k=\deg P_n$ for all $k\ge n$. So now we find from $P_k-r_kP_n\in(P_k)$ and $\deg (P_k-r_kP_n)<\deg P_n=\deg P_k$ that $P_k-r_kP_n=0$, i.e., $(P_k)=(r_kP_n)=(P_n)$.