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I am having trouble remembering how find the generators of a field.

Let's say I have $\Bbb{F}_5$ adjoined $\sqrt{2}$, and I want to show that $2+\sqrt{2}$ is a generator in F^x. So it's order is 34 and |F|=25. For some reason I was thinking I raise $2+\sqrt{2}$ to every power that is not co-prime to $24$, and if it not equal to $1$, it is a generator. But that is going to be awful, and there must be an easier way to accomplish this.

Sarah
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  • That will work and is the most direct way to do it. However, you could use the fact that the order of the field is 25, and so the member $2 + \sqrt{2}$ must have order 5 or 25 since it is not the identity element. Check 5, and if that doesn't work, you have your answer. – peabody Mar 30 '20 at 05:03
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    @peabody: The additive group is of order 25 and is isomorphic to a direct product of two groups of order 5, hence it is NOT cyclic. So your suggestion is incorrect for finding a generator for the multiplicative group. – P Vanchinathan Mar 30 '20 at 05:10
  • Sorry, @PVanchinathan I must be misunderstanding something then. I'm not certain how this group not being cyclic plays into this. If the group is of order 25, then doesn't Lagrange's theorem guarantee that the order of any element divides the order of the group? The only divisors of 25 are 1, 5, and 25, so if it's not 1 or 5 shouldn't it be 25? Can you see where my mistake is? – peabody Mar 30 '20 at 05:18
  • @peabody The question is about multiplicative order. The multiplicative group $F^*$ is cyclic of order $24$. The zero element is not in there. Because $24=3\cdot2^3$ it suffices to check that neiter $(2+\sqrt2)^8$ not $(2+\sqrt2)^{12}$ is equal to $1$. – Jyrki Lahtonen Mar 30 '20 at 05:32
  • @JyrkiLahtonen thank you, I see that now. I should have thought about the question more carefully before I posted my comment. I will be more careful next time. – peabody Mar 30 '20 at 05:37

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You know that $G=F^\times$ is a cyclic group of order $24$. Thus to show that an element $a$ is a generator, it suffices to show that $a^{12}\neq1$ and $a^8\neq 1$.

Why? Because the order of $a$ in $G$ must be a divisor of $24$. If a divisor of $24$ divides neither $12$ nor $8$, then it can only be $24$ itself.

Hence the task is simply to calculate $a^{12}$ and $a^8$.

This can be done more efficiently than multiplying $11$ times by $a$:

$$a^2=a\times a$$ $$a^4=a^2\times a^2$$ $$a^8=a^4\times a^4$$ $$a^{12}=a^8\times a^4$$

Only $4$ multiplications are needed.

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