This is wrong. Well, yes and no. Strictly speaking, this definition describes irreducibility. And irreducibility is different from primality. A prime number is actually a number, that if it divides a product, then it has to divide one of the factors. $$7\,|\,28=2\cdot 14 \;\Longrightarrow \;7\,|\,2\text{ or }7\,|\,14$$ $$4\,|\,12=2\cdot 6\text{ but }4\,\nmid \,2 \text{ and } 4\,\nmid \,6$$ It is a bit more complicated, but it is the correct definition. However, irreducible integers are prime integers and vice versa which is why the correct definition is replaced at school by the more handy one. However, there are domains in which this is not the case. The standard example is ##\mathbb{Z}[\sqrt{-5}]## where $$6=2\cdot 3=(1+\sqrt{-5})\cdot (1-\sqrt{-5})$$ is a decomposition into irreducible factors that are not prime.
