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For the $n$ distinct complex embeddings ${\sigma }_1,\dots ,{\sigma }_n$ of the number field, define the discriminant of elements of a number field byThe equality follows by multiplying the transpose of the embedding matrix by itself. Each field trace is rational: it is also the trace of the rational matrix of multiplication by $w_i{w}_j$ on $K$. Thus the determinant lies in $\mathbb{Q}$. If the $w_i$ are algebraic integers, the products and their conjugates are algebraic integers, and so are their sums. A rational algebraic integer is an integer, hence all trace entries and their determinant are integers.

Write $u_j={\sum }_i{a}_{ij}{v}_i$. The inclusion of the generated abelian groups makes $A=(a_{ij})$ an integer matrix. Its embedding matrix is the embedding matrix of the $v_i$ multiplied by $A$. Taking squared determinants givesFor independent algebraic integers $v_i$, choose an integral basis ${\omega }_i$ of ${\mathcal{O}}_K$. The same formula gives $D(v)=[{\mathcal{O}}_K:{\sum }_i\mathbb{Z}{v}_i{]}^{2}D(\omega )$. If $D(v)$ is nonzero and a square-free integer, the integer index must be one, proving that the $v_i$ form an integral basis.

Now put $\beta =3\alpha +4$. Since $\mathbb{Q}(\beta )=\mathbb{Q}(\alpha )$ has degree three, the monic polynomialis its minimal polynomial of an algebraic element. The field norm is the negative of its constant term:For the discriminant calculation, $f^{\prime }(\alpha )=3{\alpha }^2+4\alpha$. In the basis $1,\alpha ,{\alpha }^2$, using ${\alpha }^3=-2{\alpha }^2-1$, multiplication by this element has matrixThus the given derivative-norm formula gives $D(1,\alpha ,{\alpha }^2)=-59$. Since $\alpha$ satisfies a monic polynomial with integer coefficients, $1,\alpha ,{\alpha }^2$ are algebraic integers; the fact that a squarefree discriminant certifies an integral basis proves

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.