= Squarefree polynomial discriminant does not imply irreducibility
A squarefree <polynomial discriminant> does not ensure that the polynomial defines a field of its stated degree. For example, $x^3-2x-1=(x+1)(x^2-x-1)$ has <discriminant> five. Its roots have degrees one or two, so no root generates a cubic <number field>. In particular $1,\alpha,\alpha^2$ cannot be a field <basis> for any of its roots. The <squarefree discriminant certifies an integral basis> argument assumes that the polynomial is the minimal polynomial: then the <discriminant> of its power <basis> is the <field discriminant> times the square of the order index. For a reducible squarefree polynomial the analogous rank belongs instead to its quotient algebra, which is a product of fields, not to one root field.
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