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Let $A$ be an integrally closed domain, with quotient field $K_A$. My main question is the following:

Question. Does any non constant irreducible polynomial of $A[X]$ stays irreducible in $K_A[X]$ ?

Of course, this is true for monic non constant polynomials, so my first idea what to reduce to this case. This leads to the following subquestion:

Subquestion 1. Let $A$ be a domain, and let $P\in A[X]$ be an irreducible polynomial of degree $n\geq 1$, with leading coefficient $a_n$.

Is $a_n^{n-1}P(X/a_n)\in A[X]$ irreducible ? If not, is this the case if $A$ is an integral domain ?

For the moment, I've tried to prove it without any success, but I came across another subquestion, which would imply a positive answer to the previous one, and thus a positive one to the main question (if I am not mistaken...).

Subquestion 2 Let $A$ be an integral domain, and let be a non constant irreducible polynomial $P\in A[X]$. Finally , let $a\in A\setminus\{0\}$. Is is true that the divisors of $aP$ have the form , $b\in A$, or $cP$, $b,c\in A\setminus\{0\}$. If not, is is true if $A$ is an integrally closed domain ?

Greg

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    $\begingroup$ $x^2+1$ is irreducible in both $\Bbb{R}[x]$ and $\Bbb{Q}[x]$ $\endgroup$ – Abishanka Saha Jan 11 '18 at 15:25
  • $\begingroup$ Yes, but there's no connection with my question. $\endgroup$ – GreginGre Jan 11 '18 at 15:44
  • $\begingroup$ Dear @GreginGre, I think that Gauss' lemma applies here. $\endgroup$ – user90189 Jan 11 '18 at 16:46
  • $\begingroup$ Nope, $A$ is not supposed to be a GCD domain, unfortunately. $\endgroup$ – GreginGre Jan 11 '18 at 16:57
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Ok, finally the answer to the main question (and subquestion $2$) is no.

$A=\mathbb{Z}[i\sqrt{13}]$ is integrally closed, and $P=2X^2+2X+7$ is irreducible if I am not mistaken. However $2P=(2X+1+i\sqrt{13})(2X+1-i\sqrt{13})$, so in particular $P$ is not irreducible in $K_A[X]$.

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