 Solution

Since $10\equiv 2(\mathrm{mod}4)$, the ring of integers is ${\mathcal{O}}_K=\mathbb{Z}[\sqrt{10}]$, with field norm $N(x+y\sqrt{10})=x^2-10y^2$. Norms $\pm 2$ and $\pm 3$ are impossible: modulo five they would require a square congruent to 2 or 3. An algebraic integer is a unit exactly when its norm is $\pm 1$.

If 2 factored into two nonunits, their absolute norms would both be 2, since $N(2)=4$; this is impossible. The same argument for $N(3)=9$ excludes a factorization of 3. Finally $ϵ+1=4+\sqrt{10}$ has norm 6, so a nonunit factorization would require absolute norms 2 and 3, also impossible. Hence 2, 3 and $ϵ+1$ are irreducible. The conjugate $\overline{ϵ}+1$ is irreducible by the same argument.

The equality $6=2\cdot 3=(4+\sqrt{10})(4-\sqrt{10})$ gives inequivalent factorizations into irreducibles: no norm-six factor is associated to either 2 or 3, whose norms are 4 and 9. A principal ideal domain is a unique factorization domain. Consequently ${\mathcal{O}}_K$ is not a principal ideal domain.