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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 $n$.

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2 Answers 2

up vote 8 down vote accepted

$\mathbb F_{5^n}$ 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 $6$.

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Thanks! Now i can understand this properly. –  Kns May 24 '12 at 6:07
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It might also be worth mentioning that nobody knows how to do this in a way that is significantly better than brute-force enumeration. –  MJD May 24 '12 at 16:01
    
thanks for the help on the question. but it is also true that 'n' could also be any other multiple of 6, like 12, 18 and so on... all of these have the similar property, right? –  user46451 Oct 29 '12 at 16:52
    
@Ram Yes, as I said at the very end of my answer "... $\mathbb F_{5^n}$ contains an element of multiplicative order $93$ if and only of $n$ is an integer multiple of $6$." This means that $12, 18 ,24, \ldots$, all multiples of $6$, also have the same property exactly as you say. –  Dilip Sarwate Oct 30 '12 at 2:36
    
Why is "a non-trivial 93rd root of unity" an element of order 93? I'd say that any root of unity that is not 1 is non-trivial, in particular, elements of order 3 and 31 should also work. –  Phira Dec 5 '12 at 9:31

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.

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Zev - this is serious. We both need to go to sleep. –  mixedmath May 23 '12 at 8:42
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Haha neverrrrr –  Zev Chonoles May 23 '12 at 8:43
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+1 Do observe that the third cyclotomic polynomial $$\phi_3(x)=x^2+x+1$$ is quadratic, so even in characteristic zero you never need higher than a quadratic extension to include primitive cubic roots of unity! –  Jyrki Lahtonen May 23 '12 at 9:06

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