I Math Myth: A prime is only divisible by 1 and itself

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The common definition of a prime number as being only divisible by 1 and itself is misleading, as it conflates primality with irreducibility. A prime number divides a product and must also divide at least one of the factors, which is a more accurate definition. While irreducible integers are indeed prime, this relationship does not hold in all mathematical domains, such as in the ring of integers with square roots of negative numbers. An example illustrates that 6 can be factored into irreducible elements that are not prime. This nuanced understanding highlights the complexity of number theory beyond basic school definitions.
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From @fresh_42's Insight
https://www.physicsforums.com/insights/10-math-things-we-all-learnt-wrong-at-school/

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.
 
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