Bezout’s theorem fails in that ring. The particular proof given of Bezout's theorem here uses the fact that modding out the ambient ring by one of its prime (i

Bezout’s theorem fails in that ring. The particular proof given of Bezout's theorem here uses the fact that modding out the ambient ring by one of its prime (in the sense of not further factorable) elements yields a resulting ring with a prime (in the sense of not further factorable) cardinality, and therefore no nontrivial additive subgroups. However, in the ring Z[sqrt(-5)], there are irreducible elements such that modding out by them produces a ring with non-prime cardinality; for example, 2 is irreducible in this ring, but there are 4 values mod 2 in this ring. 4 is not a prime cardinal, of course. Thus, the provided proof of Bezout's Theorem can and does fail in this context. [Indeed, if we look at the range of multiplication by 2 on this particular additive group of size 4, we see that it is neither trivially of size 1 nor trivially of size all 4. It is an intermediate subgroup of size 2, which can exist because 4 has an intermediate factor of 2] The provided proof works in the integers because integers can be lined up with cardinals so that A) integer arithmetic mod n has cardinality corresponding to n, and B) also, factorizations or lack thereof for an integer are the same as for the corresponding cardinal. This ensures that integer arithmetic mod a prime integer has a prime cardinality. This fails in other contexts. [The proof of Bezout's theorem provided, incidentally, is not my favorite because it generalizes so very little. A nicer proof of Bezout's theorem, in my mind, is via the Euclidean algorithm, which then generalizes to all Euclidean domains (and, with minor modification, slightly beyond those as well, to principal ideal domains axiomatized in a manner analogous to Euclidean domains)]