Finding a generator of $\mathbb{F}_{49}^*$ 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!
 A: No, $\mathbb F_{49}$ is not the same as $\mathbb Z_{49}$. 
A: 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 $$(1+\alpha)^2 - 2(1+\alpha) - 2 = 0.$$

