Factor in and in .
Note that . If , then , so (without loss of generality) . Suppose ; then . Mod 3, we have . This is a contradiction since 2 is not a square mod 3. That is, no algebraic integer in has norm 2. (Similarly, no integer has norm 11.) Thus is irreducible as an integer in .
Evidently, . Now and . We claim that no integer in has norm 3; to that end, suppose . We must have , so that , a contradiction. In particular, both and are irreducible in .