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Let $R$ be an integral domain such that any polynomial $f(X) \in R[X]$ , which is irreducible in $R[X]$, has degree $1$. Then is it true that $R$ is a field ?

If this is not true in general , What if we also assume that $R$ is Noetherian (and normal) ?

UPDATE : The now deleted attempt to an answer sheds some light . If $0 \ne a \in R$ , then $aX^2-1$ is reducible in $R[X]$. Since this polynomial has content $1$ , we must have a factorization into one degree polynomials $aX^2-1=(cX+d)(eX+g)=ecX^2+(de+cg)X+dg$. So $dg=-1, de+cg=0, ec=a$. So $e=cg^2$, so $a=c^2g^2$ is a perfect square. So every element of $R$ is a perfect square in $R$

user
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    How about the ring of algebraic integers? – Nate Mar 07 '18 at 21:00
  • @Nate: I know that the ring of algebraic integers is integrally closed in any ring extension ... so any "monic" irreducible polynomial has degree 1 ... I don't know about non-monic ... – user Mar 07 '18 at 21:03
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    If R is Noetherian, then any element has factorization in irreducibles. But if every element of the ring is a perfect square, there are no irreducible elements in the ring, so it must be a field. – JoseCruz Mar 07 '18 at 21:10
  • @JoseCruz: Fantastic, so just the extra assumption that $R$ is factorization domain would yield $R$ is a field ... but without that assumption , the question still remains ... – user Mar 07 '18 at 21:14
  • Related: https://mathoverflow.net/questions/295012 – Watson Mar 12 '18 at 16:44

1 Answers1

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Let $K$ be an algebraically closed field and $V$ be a valuation ring of $K$. Then it is clear that $V$ is a Bezout domain. Now by Theorem 28.8 of Gilmer's book Multiplicative Ideal Theory, 1972, each irreducible polynomial $f$ over $V$ remain irreducible over $K$, and since $K$ is algebraic closed, we conclude that $f$ has degree $1$.

user26857
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Alborz Azarang
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