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 $R$
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 $1$.