6 is divisible by 2. However, "the" (really "a") prime factorization of 6 is (1+sqrt(-5)) * (1-sqrt(-5)), in which 2 does not appear.
So your assertion that "if it's divisible by a number, the number must appear in its factorization" is not true in all rings.
6 is divisible by 2. However, "the" (really "a") prime factorization of 6 is (1+sqrt(-5)) * (1-sqrt(-5)), in which 2 does not appear.
So your assertion that "if it's divisible by a number, the number must appear in its factorization" is not true in all rings.