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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 $n$?

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    The key idea is that $,P\mid Q\Rightarrow d(P) \le d(Q)\ $ & $,\ell(P),\mid \ell(Q),\ d := {\rm degree},\ \ell := \text{lead coef},,$ i.e. $,P\mid Q\Rightarrow, (d(P),,\ell (P)) < (d(Q),,\ell Q).,$ Since $N$ is well-ordered eventually the degree stablizes at its minimum, then since $R$ is well-ordered by divisibility, eventually the lead-coefs stabilize at a minimal lead coef. – Bill Dubuque Nov 14 '20 at 22:05
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    Finally by $R$ a domain: if $,0\neq P,Q\in R[x],$ have equal degree and $,P\mid Q,$ then they are associate $!\iff!$ their lead-coefs are associate (and divisibility minimal elements are associate since they must divide each other). – Bill Dubuque Nov 14 '20 at 22:06
  • @BillDubuque I can not see why $\ell(P)$ must divide $\ell(Q)$ when $P|Q$ .... could you explain this more please? –  Nov 14 '20 at 23:08
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    $\ell(PP')! =! \ell(P)\ell(P'),$ i.e. $,(a x^k! +\ldots)(bx^n! + \ldots) = ab:! x^{k+n}!+\ldots,$ and $,a,b\neq 0\Rightarrow, ab\neq 0,$ by $R$ domain. – Bill Dubuque Nov 14 '20 at 23:43
  • @BillDubuque actually my question is "Show that, if $B$ is an integral domain which satisfies the $ACCP,$ then the polynomial ring $B[t] = B^{[1]}$ also satisfies the $ACCP.$ " so I do not know how I will use the well-ordered by divisibility which eventually leads to the stabilization of the leading coefficients at the minimal leading coefficient in my case ..... do you have any suggestions? –  Nov 15 '20 at 15:31
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    Maybe try to relate AACP on $B$ to it being well-ordered by divisibility! – CPCH Nov 15 '20 at 15:53
  • @CPütz How can I do that? –  Nov 15 '20 at 15:55
  • @BillDubuque in your second comment above why you said $P,Q$ have equal degree? –  Nov 16 '20 at 05:04

1 Answers1

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  1. 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 $r_i\in R$.

  2. 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 $a_{n+i}= r_ia_{n+i+1}$.

  3. 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).

CPCH
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  • Could you please tell me what is Xam doing in the last paragraph of his answer? Is he trying to find a relation between polynomials in one unknown through relations between their coefficients? –  Nov 14 '20 at 20:43
  • In the last sentences he basically talks about why two polynomials of same degree with same leading coefficient where one is a multiple of another have to be the same (up to a unit, which does not affect the principal ideals generated by those polynomials) – CPCH Nov 14 '20 at 20:46
  • Was that what you were asking? – CPCH Nov 14 '20 at 20:46
  • Let me read your answer and everything you said and I will tell you then :) if that were what I am asking or not. –  Nov 14 '20 at 20:49
  • I do not understand your last part of line 3, why $$ \deg P_{n+i+1}=\deg R_i + \deg P_{n+i+1}$$? What is $R_{i}$? did you mean $r_{i}$ instead? –  Nov 14 '20 at 23:52
  • In that third line, how did you use the fact that $R$ is an integral domain? Is that definition of the degree function you used is only correct for integral domains? –  Nov 14 '20 at 23:57
  • I think in 2, the second line you meant $P_{n + i}$ ..... am I correct? –  Nov 15 '20 at 02:13
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    Yep, edited that. The degree is only additive on integral domains. Otherwise there would be zero divisors on $R$, for example $a,b\neq 0$ with $ab=0$ and $\deg aX+\deg bX\neq\deg aX\cdot bX=\deg 0X^2$. – CPCH Nov 15 '20 at 09:59
  • And what is the meaning of the symbol $\sim$ in Xam's answer in the link? –  Nov 15 '20 at 10:53
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    ‚Associated‘ as in ‚generating the same principal ideal‘ – CPCH Nov 15 '20 at 11:02
  • I think also in your fourth line you meant "here we used that $R[X]$ is an integral domain" not R ...... am I correct? –  Nov 15 '20 at 15:22
  • Let us continue this discussion in chat. – CPCH Nov 15 '20 at 15:35
  • Why we need to differentiate between the deg. of the polynomial being stationary and the leading coefficient being stationary? –  Nov 15 '20 at 17:40