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$f$ is said to be primitive if $(a_0, a_1, ..., a_n)=(1)$ Prove that if $f, g \in A[x]$ then $fg$ is primitive $\iff f, g$ primitive

I have read the answers to this question here: Exercise from Atiyah-Macdonald, Chapter 1, 2.iv)

but I need more details to understand the proof. One of the comments says to prove it for $\deg(f)=\deg(g)=1.$

EDIT: Why is this question closed? Question 2. isnt answered in the link above - as the private message says. And the question has two upvotes!

$\implies.$ $fg=(a_0+a_1x)(b_0+b_1x) = (a_0b_0)+(a_1b_0+a_0b_1)x + (a_1b_1)x^2$ and we have $(a_0a_1)=(1)$, and $(b_0b_1)=(1)$

  1. What is the ideal $(a_0, a_1)$? I have seen the ideal $(a_0)(a_1)$, $(a_0)+(a_1)$. Edit: answere: (r,s)={rx+sy|x,y∈R} = (r+s)
  2. How do we show $fg = 1$ when both have degree $1$ ?
  3. How do we show the inductive step when $\deg(f)=n$, $\deg(g)=m$?
Endre Moen
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    See also https://math.stackexchange.com/questions/413788/product-of-two-primitive-polynomials . – darij grinberg Sep 03 '20 at 09:33
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    Does this answer your question? Exercise from Atiyah-Macdonald, Chapter 1, 2.iv) – Quillo Sep 03 '20 at 09:46
  • For (1): as you note in the statement of your question, it is common to write $(a_0, ..., a_n)$ to be the ideal generated by $a_0, ..., a_n$. For (2): If you want to do induction, you'll only need to start at the degree $0$ case, and the rest is covered by the second answer to the question you linked - it may be helpful for you to specialise that answer to the degree 1 case. – Mummy the turkey Sep 03 '20 at 10:42
  • Q1: I found the answer: (r,s)={rx+sy|x,y∈R} = (r+s) – Endre Moen Sep 03 '20 at 10:55
  • Q2: Is it $(a_0)(a_1)(b_0)(b_1)(fg) = (a_0)(a_0b_0)(a_1)(b_0)(b_1) + ... +(a_0)(a_1b_1)(a_1)(b_0)(b_1)$ = $(1) (a_0b_0) + (1)(a_1b_0+a_0b_1)x+ (1)(a_1b_1)x^2 = (1)$ ? – Endre Moen Sep 03 '20 at 11:00
  • @Quillo - One of the comments says to prove it for deg(f)=deg(g)=1 - is helpful, but I dont fully understand the proof, no. – Endre Moen Sep 03 '20 at 11:05
  • It is false that $fg = 1$ when $\deg f = \deg g = 1$. Not sure where you got this from. Recall the notation for ideals generated by a list of elements in a commutative ring: $\left(a_1, a_2, \ldots, a_k\right) = \left{r_1a_1 + r_2a_2 + \cdots + r_ka_k \mid r_1, r_2, \ldots, r_k \in R\right}$ where $R$ is the ring. Thus, in particular, $\left(a\right) = \left{ra \mid r \in R\right}$. Note that $\left(a_1, a_2, \ldots, a_k\right) = \left(a_1\right) + \left(a_2\right) + \cdots + \left(a_k\right)$, where the $+$ signs on the right hand side stand for addition of ideals. – darij grinberg Sep 03 '20 at 20:24
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    I have edited the answer at https://math.stackexchange.com/a/824086/ to make it somewhat clearer, but it is not the epitome of easy reading. You might want to read https://arxiv.org/abs/1209.6307 instead (Sections 1-2 should suffice). – darij grinberg Sep 03 '20 at 20:27

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