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Is this true?

In every integral domain, irreducible implies prime

No — it is false.

The counterexample

2 in $Z [\sqrt - 5]$

2 is irreducible (no element of norm 2) but divides $(1 + \sqrt - 5) (1 - \sqrt - 5) = 6$ without dividing either factor.

The kind of mistake this is

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Others that fail the same way

From Rings and Fields › Euclidean, PID, UFD hierarchy

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