Primitive Roots of Unity and THE primtive root of unity

Question

If a question states that $\omega_n$ is THE primitive complex root of unity of order $n$, then does this always imply that $\omega_n$ = $e^{2\pi i/n}$, i.e. primitive roots of unity of order $n$ are those complex numbers such that $\omega_n = e^{2\pi ij/n}$ for which $j$ is such that $0 < j < n$ and $gcd(j,n)$= 1, but if a question asks for THE primitive root, can it be taken as the case $j=1$?

Answer 1

The equation amounts to solutions for $\omega^n=1$.

The solution for this is $\omega=\exp{(2\pi i j/n)}$.

The solution is primitive when the $\gcd(j,n)=1$, which means that the first power to equal 1, is when n divides the power.

The solution is THE primitive exactly when $j=1$.

The reason for this, is that isomorphisms are used to project the primitive (which is taken as the identity in isomorphisms), onto other solutions. The same situation occurs in

$x^2=4$. Thus, $2$ and $-2$ are square roots of $4$, but $x=2$ is THE square root of $4$.

Answer 2

As explained in the comments to the question, mathematicians do not in general speak of "the" primitive $n$th root of unity because all such roots behave isomorphically.

There are cases where choosing different primitive roots can give disagreeing results because of some outside convention not inherent to the roots. For instance, we can use the theory of quadratic Gauss sums to render

$\zeta-\zeta^2-\zeta^3+\zeta^4 = \sqrt5,$

where $\zeta$ is a primitive fifth root of unity; but if we apply the usual definition that the square root should be nonnegative then the relation holds only if we choose a $\zeta$ candidate with a positive real part. But even here that admits a pair of complex-conjugate choices, so none emerges as deterministically "the" primitive fifth root of unity with respect to the Gauss-sum relation.