Finding the values of $n$ for which $\mathbb{F}_{5^{n}}$, the finite field with $5^{n}$ elements, contains a non-trivial $93$rd root of unity

Question

For which values of $n$, does the finite field $\mathbb{F}_{5^{n}}$ with $5^{n}$ elements contain a non-trivial $93$rd root of unity?

I don't know how to find the value of $n$.

Answer 1

$\mathbb F_{5^n}$ contains an element of multiplicative order $93$ if and only if $93$ is a divisor of $5^n-1$, that is, $5^n \equiv 1\bmod 93$. The brute force way of finding the smallest value of $n$ is to just calculate the values of $5^1$, $5^2$, $5^3$, $\ldots$ modulo $93$ until you find the answer. So we proceed as follows: $$\begin{align*} 5^1 &\equiv 5 \bmod 93\\ 5^2 &\equiv 25 \bmod 93\\ 5^3 = 125 &\equiv 32 \bmod 93\\ 5^4 \equiv 5\times 32 =160 &\equiv 67 \bmod 93\\ 5^5 \equiv 5\times 67 = 335 &\equiv 56 \bmod 93\\ 5^6 \equiv 5\times 56 = 280 &\equiv 1 \bmod 93\\ \end{align*}$$ Thus, $\mathbb F_{5^6}$ is the smallest field of characteristic $5$ that contains an element of multiplicative order $93$. Since $5^6-1$ is a divisor of $5^n - 1$ if and only if $6$ is a divisor of $n$, we conclude that $\mathbb F_{5^n}$ contains an element of multiplicative order $93$ if and only if $n$ is an integer multiple of $6$.

Answer 2

I'll give you one direction - see if you can do the other on your own.

Suppose we have a non-trivial $93$rd root of unity $\zeta\in\mathbb{F}_{5^n}$. Then $\zeta$ would, of course, have to be non-zero. So $\zeta\in\mathbb{F}_{5^n}^\times$, and by definition we have that the order of $\zeta$ as an element of the multiplicative group $\mathbb{F}_{5^n}^\times$ is $93$. By Lagrange's theorem, this is only possible if $93\mid 5^n-1$. Note that $93\mid 5^n-1$ if and only if $3\mid 5^n-1$ and $31\mid 5^n-1$, because $93=3\cdot 31$ is the prime factorization of $93$.

When does that happen? Look at the following table: $$\begin{array}{c|c|c|c|c|c|c} n & 0 & 1 & 2 & 3 & 4 & 5 & 6\\ \hline 5^n & 1 & 5 & 25 & 125 & 625 & 3125 & 15625\\ \hline \text{mod }3 & 1 & 2 & 1 & 2 & 1 & 2 & 1 \\ \hline \text{mod }31 & 1 & 5 & 25 & 1 & 5 & 25 & 1 \end{array}$$ The period of $5^n$ modulo $3$ is $2$, and the period of $5^n$ modulo $31$ is $3$ (the proof that this period holds for all $n$ is straightforward). Thus, the $n$ for which $5^n\equiv 1\bmod 3$ and $5^n\equiv 1\bmod 31$ (i.e. the $n$ for which $3\mid 5^n-1$ and $31\mid 5^n-1$) are those $n$ such that $2\mid n$ and $3\mid n$, i.e. the $n$ which are multiples of $6$.

Ok, so we've established that, if there is a primitive $93$rd root of unity in $\mathbb{F}_{5^n}$, then it must be the case that $6\mid n$.

Is the converse true? Try to work it out yourself.