Let \(K\) and \(K'\) be number fields. For a rational prime \(p\) and \(\ell\geq 1\), let \(k_\ell(p)\) be the number of prime factors of \(p\) in \(K\) that have inertial degree equal to \(\ell\), and define \(k_\ell'(p)\) to be the corresponding value for \(K'\). Let \(F=\sum^\infty_{\ell=1} a_\ell X_\ell\) be an infinite-dimensional linear form, where the coefficients \(a_1,a_2,\dots\) are integers. For any prime \(p\), set \(F_K(p)=F(k_1(p), k_2(p),\dots)\) and \(F_{K'} (p)=F(k_1'(p), k_2'(p),\dots)\). Then \(K\) and \(K'\) are called \(F\)-linearly equivalent if, for almost all \(p\), one has \(F_K(p)=F_{K'}(p)\). Furthermore, for an infinite-dimensional linear form \(F\) as above, consider the condition \[ \sum_{d\mid m}\mu(d){m\over d}a_d\neq 0\text{ for all }m=1,\dots, c. \tag{1} \] The main theorem reads as follows: Theorem. Fix and integer \(c\geq 2\). Two algebraic number fields \(K\) and \(K'\) of degrees at most \(c\) over the rationals are arithmetically equivalent if and only if they are \(F\)-linearly equivalent through a linear form \(F\) that satisfies (1).