Finding a generator of $\mathbb{F}_{49}^*$

Question

I have a little doubt.

Let $\alpha\in\mathbb{F}_{49}$ such that $\alpha^2=3$ and $\mathbb{F}_{49}=\mathbb{F}_7(\alpha)$. Find a generator for the cyclic group $\mathbb{F}_{49}^*$ and find the minimal polynomial over $\mathbb{F}_7$

For the first part, it's enough to find a primitive root modulo 49?

Thanks!

Answer 1

This are the powers of $1 + \alpha$ in $\mathbb{F}_{49}$:

$$\begin{array}{cccccc} \color{blue}{1 + 1\alpha} & 4 + 2\alpha & 3 + 6\alpha & 0 + 2\alpha & \color{blue}{6 + 2\alpha} & 5 + 1\alpha \\ \color{blue}{1 + 6\alpha} & \color{red}{5 + 0\alpha} & 5 + 5\alpha & 6 + 3\alpha & \color{blue}{1 + 2\alpha} & 0 + 3\alpha \\ \color{blue}{2 + 3\alpha} & 4 + 5\alpha & 5 + 2\alpha & \color{red}{4 + 0\alpha} & \color{blue}{4 + 4\alpha} & 2 + 1\alpha \\ \color{blue}{5 + 3\alpha} & 0 + 1\alpha & 3 + 1\alpha & 6 + 4\alpha & \color{blue}{4 + 3\alpha} & \color{red}{6 + 0\alpha} \\ \color{blue}{6 + 6\alpha} & 3 + 5\alpha & 4 + 1\alpha & 0 + 5\alpha & \color{blue}{1 + 5\alpha} & 2 + 6\alpha \\ \color{blue}{6 + 1\alpha} & \color{red}{2 + 0\alpha} & 2 + 2\alpha & 1 + 4\alpha & \color{blue}{6 + 5\alpha} & 0 + 4\alpha \\ \color{blue}{5 + 4\alpha} & 3 + 2\alpha & 2 + 5\alpha & \color{red}{3 + 0\alpha} & \color{blue}{3 + 3\alpha} & 5 + 6\alpha \\ \color{blue}{2 + 4\alpha} & 0 + 6\alpha & 4 + 6\alpha & 1 + 3\alpha & \color{blue}{3 + 4\alpha} & \color{red}{1 + 0\alpha} \end{array}$$

(in $\color{red}{\text{red}}$ all the elements in $\mathbb{F}_7$ and in $\color{blue}{\text{blue}}$ all the generators).

We know that $\mathbb{F}_{49}^*$ has $48$ elements and if $g$ is a generator, then $g^k$ is also a generator for every positive integer $k < 48$ coprime with $48$. Thus, there are $16$ generators. We can discard the elements in $\mathbb{F}_7^*$ and the elements in $\alpha\mathbb{F}_7^*$ since those cannot have degree larger than $6$ and $12$, respectively. Therefore we have $16$ generators out of $36$ candidates. I suppose one could use some more powerful tools to narrow even more the search, but I was lucky with the first pick.

Its minimal polynomial over $\mathbb{F}_7$ is $X^2 - 2X - 2$.

Indeed it has to be of degree $2$ since $\mathbb{F}_7(\alpha)$ is a quadratric extension over $\mathbb{F}_7$. And $$(1+\alpha)^2 - 2(1+\alpha) - 2 = 0.$$

Answer 2

No, $\mathbb F_{49}$ is not the same as $\mathbb Z_{49}$.