Integral domain over which any non-constant, one variable, irreducible polynomial has degree 1

Question

Let $R$ be an integral domain such that every non-constant, irreducible polynomial $f(X) \in R[X]$ has degree $1$.

Q. is it true that $R$ is a field?

If $0 \ne a \in R$ , then $X^2-a$ is reducible in $R[X]$. Since this polynomial has content $1$ , we must have a factorization into one degree polynomials $X^2-a=(cX+d)(eX+g)=ecX^2+(de+cg)X+dg$. So $ec=1, de+cg=0, dg=-a$. So $d=-c^2g$, hence $a=-dg=(cg)^2$ . So every element of $R$ is a perfect square in $R$ . So if $R$ is a factorization domain with only irreducible non-constant polynomials in $R[X]$ being of degree $1$, then $R$ has no irreducible elements , hence $R$ must be a field.

I have no idea about what happens if $R$ is not a factorization domain.

Answer 1

The answer is no, as shown by the following example.

Take as $R$ the ring $\bar{\mathbb{Z}}$ of algebraic integers. Since its fraction field is $\bar{\mathbb{Q}}$, which is algebraically closed, it follows by Gauss lemma for GCD domains that the only non-constant, irreducible polynomials in $\bar{\mathbb{Z}}[x]$ have degree $1$.

Remark. The ring $\bar{\mathbb{Z}}$ is non-Noetherian, for instance because there is an infinite ascending chain of (principal) ideals $$(2) \subset (2^{1/2}) \subset (2^{1/4}) \subset \ldots $$

Since it is a Bézout domain, it follows that it is not a UFD; in fact, for a Bézout domain, being Noetherian, PID and UFD are three equivalent conditions.

Answer 2

Let K be an algebraic closed field and V be a valuation of K, then it is clear that V is a Bezout domain. Now by Theorem 28.8 of Gilmer book "multiplicative ideal theory" 1972, each irreducible polynomial f(x) over V remain irreducible over K, and since K is algebraic closed, we conclude that f has degree 1.