Gauss lemma for Irreducibility over $\mathbb{Z}\bigl[\sqrt{13}\mkern2mu\bigr]$ Gauss lemma states that a monic polynomial over $\mathbb{Z}[x]$ is irreducibile iff it is Irreducibile over $\mathbb{Q}[x].$
My concern is whether this statement is valid for monic polynomials over $\mathbb{Z}\bigl[\sqrt{13}\mkern2mu\bigr]$ and $\mathbb{Q}\bigl[\sqrt{13}\mkern2mu\bigr].$
 A: A version of Gauss's lemma is true in far more generality. Here's the most general version I know: If $A \subset B$ is an inclusion of integral domains and $K$ is the field of fractions of $A$, then $B \otimes_A K$ is an integral domain. (One can prove this by checking that $B \otimes_A K$ is isomorphic to the localization of $B$ at the multiplicative subset $A \setminus \{0\}$, so $B \otimes_A K$ is naturally a subring of the field of fractions of $B$.)
Here's how we relate this back to polynomials: Let $R$ be an integral domain with field of fractions $F$, and let $g \in R[x]$ be a monic prime element. We have an injective homomorphism $R \to R[x]/(g)$, so applying the above theorem, we conclude that $R[x]/(g) \otimes_R F \cong F[x]/(g)$ is an integral domain, hence $g$ is prime in $F[x]$.
If $R$ happens to be a UFD, then "prime" is equivalent to "irreducible", so we have the familiar statement about irreducible polynomials. In general, the analogous statement holds for prime polynomials. So you don't need to check whether the ring is a UFD, you just need to check whether the irreducible polynomial in question is prime.
