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Find a primitive element $\alpha$ in $\mathbb{F}_{25}$ and for every $\beta \in \mathbb{F}_{25}^*$ find the least $n\in \mathbb{Z}^+$ such that $\alpha^n=\beta$.

I constructed $\mathbb{F}_{25}$ by $\mathbb{F}_{5} / (x^2+2x+3)$ but I am not sure how to find a primitive element as there are $25$ orders to compute. I tried $\alpha$ as a root of the polynomial I used, and I got $\alpha^3=1$ so obviously that is not a generator $\mathbb{F}_{25}^*$. I found another construction by $\mathbb{F}_{5} / (x^2+4x+2)$ and then the root of this polynomial, say $\alpha$, in $\mathbb{F}_{5} / (x^2+2x+3)$ is primitive but I would like to know how to do this for my construction.

mandella
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  • You have to calculate $a^i, \forall i \in {0,1,2,\ldots , 24}$ and use the denominator to simplify each one i.e. $\alpha^2 + 2\alpha + 3 = 0$ because it's a quotient field. – K.K.McDonald Jan 07 '20 at 09:51
  • In an old answer of mine I show how to find roots of unity of orders eight and three in $\Bbb{F}_{25}=\Bbb{F}_5[\sqrt2]$. That gives an alternative way that I find kinda cute. – Jyrki Lahtonen Jan 07 '20 at 10:13

2 Answers2

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When you say $\Bbb F_5/(x^2+2x+3)$, I think you ought to write $\Bbb F_5[x]/(x^2+2x+3)$. Because that's what I think you mean.

If $\alpha$ is a root of your polynomial, then we have $\alpha^2 = 3\alpha + 2$, and $$\alpha^3 = \alpha^2\cdot \alpha = (3\alpha +2)\alpha\\ = 3\alpha^2 +2\alpha = 3(3\alpha + 2) + 2\alpha\\ = \alpha + 1\neq 1$$ Here is a full table of the powers of $\alpha$, generated by a rather simple Python script (the print format may look strange, but it's specifically constructed to make it copy-pastable into the below table, with minimal need for tidying up): $$ \begin{array}{|c|c|} \hline n & \alpha^n\\ \hline 0 & 1\\ 1 & \alpha\\ 2 & 3\alpha + 2\\ 3 & \alpha + 1\\ 4 & 4\alpha + 2\\ 5 & 4\alpha + 3\\ 6 & 3\\ 7 & 3\alpha\\ 8 & 4\alpha + 1\\ 9 & 3\alpha + 3\\ 10 & 2\alpha + 1\\ 11 & 2\alpha + 4\\ 12 & 4\\ 13 & 4\alpha\\ 14 & 2\alpha + 3\\ 15 & 4\alpha + 4\\ 16 & \alpha + 3\\ 17 & \alpha + 2\\ 18 & 2\\ 19 & 2\alpha\\ 20 & \alpha + 4\\ 21 & 2\alpha + 2\\ 22 & 3\alpha + 4\\ 23 & 3\alpha + 1\\ \hline \end{array} $$

Arthur
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    Oh my god, when I calculated $\alpha^3$ I did $3\cdot3=6$ so that is why I got $1$ at the end. – mandella Jan 07 '20 at 09:55
  • So the second part of the question is actually trivial, as $\alpha$ is primitive so each element is already a power of $\alpha$, right? – mandella Jan 07 '20 at 09:58
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    @mandella Sure, but it's not trivial to figure out which power of $\alpha$. And it's not entirely trivial that $\alpha$ is primitive either. For instance, a root of the polynomial $x^2 + 2$ (which is to say, $\alpha + 1$ or $4\alpha + 4$) will not be primitive, as those roots generate only every third element in this list. So you could've been unlucky with which polynomial you chose in the first place. – Arthur Jan 07 '20 at 10:02
  • Yes, I understand that, but I guess if I could form the table that you did, then it would be trivial. – mandella Jan 07 '20 at 10:04
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    @mandella I think that, in this specific case, just generating the table is the most efficient thing to do, at least if you have some scripting available to do all the arithmetic for you (so you avoid mistakes like $3\cdot 3 = 6$; I did such a mistake myself when I got to $\alpha^6$ trying to make the table by hand before giving up and writing a script). If it turns out you chose an $\alpha$ that wasn't suitable, just go with $\alpha +1$, and try again, and if that fails, $\alpha + 2$. Eventually you must necessarily find a primitive element. – Arthur Jan 07 '20 at 10:07
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If $\alpha \in \mathbb F_{25}$ has minimal polynomial $P$ over $\mathbb F_5$, then $\alpha$ is a primitive element iff $P$ divides the $24$th cyclotomic polynomial over $\mathbb F_5$.

So to prove that $\alpha$ is a primitive element, it suffices to check that $x^2+2x+3$ does not divide $\Phi_1 = x-1$, $\Phi_2=x+1$, $\Phi_3 = x^2+x+1$, $\Phi_4 = x^2+1$, $\Phi_6 = x^2+4x+1$, $\Phi_8 = x^4+1 = (x^2+2)(x^2+3)$, $\Phi_{12} = x^8-x^4+1 = (x^2+2x+4)(x^2+3x+4)$.

Bart Michels
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  • This is more or less equivalent to the argument "$\Bbb F_{25}^*$ is a cyclic group of order $24$, so it suffices to check that $\alpha, \alpha^2, \alpha^3, \alpha^4, \alpha^6, \alpha^8$ and $\alpha^{12}$ are all different from $1$", isn't it? – Arthur Jan 07 '20 at 10:18
  • Yes, but computationally easier to check, for a human. A computer might prefer your answer over factoring polynomials. – Bart Michels Jan 07 '20 at 10:21
  • Not necessarily. You can check those powers of $\alpha$ pretty quickly by repeated squaring or similar (with the constant risk of miscalculation, of course, and not miscalculation which is easy to catch), but for your approach, one would have to go about remembering all the different cyclotomic polynomials. Not saying my approach is faster or better, but I don't find it clear-cut. Also, I would say my approach is... more approachable, with less theory. – Arthur Jan 07 '20 at 10:28