# If an element of a field is a primitive $7$th root, why are its powers also $7$th roots?

If $F$ is a field containing an element $\alpha$ that is a primitive seventh root of $1$, why is it true that any power $\alpha^k$ is also a seventh root of $1$? Also, why is the set $\{1, \alpha, \alpha^2...\alpha^6\}$ the full set of seventh roots in $F$?

• Are you familiar with the result that a degree $n$ polynomial has at most $n$ roots in a field? – hardmath Jul 28 '16 at 20:33
• To see whether $\alpha^k$ is a 7th root of $1$, just try computing $(\alpha^k)^7$. – Eric Wofsey Jul 28 '16 at 20:34
• Because $(a^k)^7=(a^7)^k=1^k=1$. – fleablood Jul 28 '16 at 20:36

If $\alpha \in F$ is a primitive seventh root of $1$, then it generates a cyclic subgroup of order $7$ inside $F$ consisting of $\{1, \alpha, \alpha^{2}, \ldots, \alpha^{6}\}$. By Lagrange, each of these elements has order dividing $7$, and since only $1$ has multiplicative order $1$, $\alpha, \alpha^{2}, \ldots, \alpha^{6}$ must each have order $7$, and therefore are also (primitive) seventh roots of $1$. In particular, each $\alpha^{k}$ for $k = 0, 1, \ldots, 6$ is distinct and is a root of $X^{7}-1$ over $F$. Since $F$ is a field, $X^{7} - 1$ has at most $7$ roots in $F$, so these powers of $\alpha$ must be all such roots.
Let $F$ be a field. Every polynomial in $F[X]$ can be decomposed in linear factors (over some algebraic closure of $F$) and has therefore a number of roots in $F$ that does not exceed its degree. In particular, the polynomial $X^7=1$, whose solutions are exactly the seventh roots of unity, has at most 7 solutions. If you have found a non-trivial root $\alpha$, then you have $(\alpha^k)^7=(\alpha^7)^k=1$ for all $k$. The fact that $\alpha$ is primitive means exactly that it generates the whole group of seventh roots of unity.
More generally, if $\alpha$ is an $n$-th root of $1$, then every power of $\alpha$ is again an $n$-th root, because $$(\alpha^k)^n=(\alpha^n)^k=1^k=1$$
What's true in the case of a primitive $n$-th root is that the powers $1=\alpha^0, \alpha=\alpha^1, \alpha^2,\dots,\alpha^{n-1}$ are all the $n$-th roots.
In the case of $n=7$, since it is prime, every power $\alpha^k$, with $0<k<7$ is again a primitive $7$-th root, because the set of $7$-th roots is a simple cyclic group, so every non-identity element is a generator.