Return to Answer

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 (for a Bézout domain, being Noetherian, PID and UFD are equivalent conditions).

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.

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 (for a Bézout domain, being Noetherian, PID and UFD are equivalent conditions).

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 (for a Bézout domain, being Noetherian, PID and UFD are equivalent conditions).

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 $1$.

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 (for a Bézout domain, being Noetherian, PID and UFD are equivalent conditions).