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Let $R$ be an integral domain and $F$ be its field of fraction. We know that if $R$ is a UFD then for $f(X), g(X) \in R[X],$ $f(X)g(X)$ is primitive iff $f(X)$ and $g(X)$ are primitive. BTW by $f(X)$ primitive we mean $C(f)$=$\text{gcd}$ of all non zero coefficiens of $f(X)$ in $R$ is $1.$

Now my question is the following: Does it hold if we are given that $R$ is integrally closed.

Actually I want to show that if a monic polynomial $f(X) \in R[X]$ is irreducible in $F[X]$ then it is also irreducible in $R[X],$ where $R$ is integrally closed. Can I show that even if the above doesn't hold. I need some help. Thanks.

user371231
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  • See the following extensive survey GCD Domains, Gauss’ Lemma, and Contents of Polynomials by D. D Anderson. In particular see the results around superprimitive polynomials (dating back to Hwa Tsang Tang's thesis under Kaplansky in the early 70's). That will answer your questions and then some. – Bill Dubuque Oct 28 '18 at 16:30
  • What precisely is "the other way" that you seek? The direction you state it is trivial, since any nontrivial factorization over $R$ persists in its fraction field. – Bill Dubuque Oct 28 '18 at 16:49
  • Is it trivial ? I don’t know how it is trivial. Reducibility of $f \in R[X$ imply that $f=gh$. Now it may happen that g is non zero constant in $R$. How do I complete after that ? – user371231 Oct 28 '18 at 17:13
  • In your question you said $f$ is monic so $f=gh,,g\in R,\Rightarrow, g$ divides the lead coef of $f$ so $g$ is a unit. If you follow the links I gave you will learn that Tang showed that $R$ being integrally closed is equivalent to this: $ $ if $f\in R[x],, g\in F[x],$ then $, fg\in R[x]\iff f g_i \in R[x],$ for all coefs $,g_i,$ of $g$. This leads to the notion of superprimitive polynomials, and the study of PSP domains where primitive $\Rightarrow$ superprimitive and related notions, all of which are discussed at length in said survey – Bill Dubuque Oct 28 '18 at 17:38

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No, GL = Gauss's Lemma is not true for any integrally closed domain because it implies that atoms (irreducibles) are prime. Indeed, if an irreducible $\,p\,$ is not prime then there exists $a,b\,$ such that$\,p\mid ab,\, p\nmid a,p\nmid b\,$ so $\,f= px+a,\,g = px+b\,$ are primitive but $\,p\mid fg\,$ so $fg$ is not, so GL fails. So a counterexample lies in any integrally closed domain having a non-prime irreducible (e.g. any non-UFD quadratic number ring).

To learn more anout generalizations of Gauss's Lemma I highly recommend the following extensive survey GCD Domains, Gauss’ Lemma, and Contents of Polynomials by D. D Anderson (free link).

In particular see the results around superprimitive polynomials (dating back to Hwa Tsang Tang's thesis under Kaplansky in the early 70's). Tang showed that $R$ being integrally closed is equivalent to this: $ $ if $\,f\in R[x],\,g\in F[x]\,$ then $\,fg\in R[x]\iff fg_i\in R[x]\,$ for all coefs $\,g_i$ of $g.\,$ This naturally leads to the notion of superprimitive polynomials, and the study of so-called PSP domains where primitive $\Rightarrow$ superprimitive and related notions, all discussed at length in said survey.

Bill Dubuque
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  • But in my eye this is not that what I really want to prove. What you have show implies that if a monic polynomial $f \in R[X]$ is irreducible in $R[X]$ then it is also irreducible in $F[X]$. I am looking for the other way. – user371231 Oct 28 '18 at 16:21
  • @user371231 I have edited the answer. – Bill Dubuque Oct 28 '18 at 19:02
  • Please clarify what you mean by a primitive polynomial in a non-GCD domain. – user26857 Oct 29 '18 at 06:47
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    @user26857 The definition used above is $f$ is primitive if every common divisor of its coefficients is a unit. – Bill Dubuque Oct 29 '18 at 17:37
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Being integrally closed is equivalent to having irreducibility of monic polynomials in $R[x]$ persist in $F[x]$.

Check out theorem 3 of this short paper, to which Anderson also contributed.

The paper also shows that if you drop the word 'monic' above then you get precisely the concept of a Schreier domain, i.e. an integrally domain such that $x \mid ab$ implies $x = x_1 x_2$ and $x_1 \mid a, x_2 \mid b$.

One of the tools used in their proofs is a nice formula for the contents of a product of polynomials over integrally closed domains, though understanding it requires some familiarity with the concept of divisorial ideals. Letting $c(h)$ denote the ideal generated by the coefficients of the polynomial $h$ and $I_v$ denote the divisorial closure of the ideal $I$, we have that

$$\big[c(f)c(g)\big]_v = c(fg)_v$$ over any integrally closed domain.

Badam Baplan
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  • Yes, that also follows quickly from the older results of Tang that I mentioned aboce (they cite that paper). – Bill Dubuque Oct 29 '18 at 01:34
  • Right, I just wanted to supplement that part of your answer with a shorter and more directly relevant paper, since the survey on GCD domains is a lot of material to digest.

    I also thought it worth mentioning the content formula perspective, which is entirely absent in Tang's important but dated work. Almost all of the results in that paper are trivial to deduce from Dedekind-Mertens.

     – Badam Baplan Oct 29 '18 at 01:50
  • Iirc there are further results in Tang's thesis (I no longer have a copy to check). But I do agree that more recent paper is well-worth reading too, esp. for the refinement-based (Schreier) perspective. And +1. – Bill Dubuque Oct 29 '18 at 02:15