How to find generators of the multiplicative group of $\mathbb{F}_9$

Question

Find generators of the multiplicative group of $\mathbb{F}_9$

I construct $\mathbb{F}_9=\frac{\mathbb{Z}_3[x]}{\langle x^2-2 \rangle}$

I know elements in $\mathbb{F}_9$ take the form $ax+b$ where $a,b \in \mathbb{Z}_3$.

Then how to find all generators for the multiplicative group of $\mathbb{F}_9$?

Answer 1

Since the multiplicative group has order $8$, every element has order $1,2,4,8$. Therefore, any non-generator has order $1,2,4$ and hence is a solution to $$Y^4=1$$ You know that $1,2$ are the two solutions to $Y^2-1=0$ and if you figure the two solutions to $Y^2+1=0$, by elimination, everything else is a generator.

By construction $$X^2+1=X^2-2 =0$$ therefore $Y= \pm X$ are the two elements of order $4$.

Answer 2

There are only eight elements in the multiplicative group of $\mathbb{F}_9$, and since the group is isomorphic to $\mathbb{Z}/8\mathbb{Z}$ which has four generators, just guessing some element and checking if it's a generator isn't too bad of an idea. You only have to verify that your guess doesn't have order two or four, and you are done.

In the quotient where $\mathbb{F}_9$ is constructed as $\mathbb{F}_3[x]/\langle x^2-2 \rangle$, I've got that the element corresponding to $x+2$ generates the whole group because it has order greater than four. $$\begin{align} \\(x+2)^1 = \;&x+2 \\(x+2)^2 = \;&x \\(x+2)^3 = \;&2x+2 \\(x+2)^4 = \;&2 \\(x+2)^5 = \;&\dotsb \end{align}$$

And you can find the other generators through trial and error similarly.

Answer 3

Here's what you can also do.

Because there are eight non-zero elements in $\Bbb{F}_9$ you are looking for an eighth root of unity. Let's draw a parallel to the case of complex numbers, and see what's cooking.

In $\Bbb{C}^*$ an eighth root of unity is given by $$ \zeta_8=e^{\pi i/4}=\cos(\pi/4)+i\sin(\pi/4)=\frac{1+i}{\sqrt2}. $$ So it makes sense to try and identify substitutes for $i$ and $\sqrt2$ in your field.

Because $x^2-2=x^2+1$, we know that $u:=x+\langle x^2-2\rangle$ can serve in the role of $i$. After all, $u^2=-1$. Wait a second! Because $-1=2$, $u$ can also serve in the role of $\sqrt2$! So this is plain sailing! We get $$ \zeta_8=\frac{1+u}u=\frac1u+1 $$ as a candidate. As $u^4=1$, we see that $u^3=u^2\cdot u=-u$ is the reciprocal $1/u$.

So $$\zeta_8=-u+1$$ should work. Let's check: $$ \zeta_8^2=(1-u)^2=1-2u+u^2=(1+u^2)-2u=(1-1)+u=u. $$ As we already know that $u$ is of order four, we can conclude that our $\zeta_8$ is of order eight and hence a generator.

Altogether there are $\phi(8)=4$ generators. You get the others by using a different subsitute for $\sqrt{-1}$ and $\sqrt2$ in the above calculation. Just pick any sign combination in $i=\pm u$ and $\sqrt2=\pm u$, and the formula $\zeta_8=(1+i)/\sqrt2$ will give you a generator.

Answer 4

Perhaps exactly the same answer as everybody else’s, but in different words:

Since $2\equiv-1\pmod3$, you’re adjoining $i$ to $\Bbb F_3$. You know that $\{\pm1,\pm i\}$ aren’t generators, so use $1+i$ as a generator, and let’s call it $g$. Direct hand calculation shows that $g^3=1-i$, $g^5=-1-i$, and $g^7=2+i$. These four are the generators of $\Bbb F_9^\times$.