Algebra, finding the elements of the field and solving irreducible polynomials I'm trying to do this problem from a practice final but there are no solutions.
I honestly am pretty stumped.
My thought was since it has 7 elements, then the degree of the polynomial must be one and we must be working with the field $F_7$
1a) I have no idea what that means, would the squares and cubes just be $0^2, 1^2, 2^2, 3^2 ... , 7^2$ and $0^3, ... ,  7^3$?
1b) My thought is, the field F is 7, so L must be a field with a polynomial of degree 2, but can it just be any polynomial? Don't know how to solve the polynomials $x^2-3$ and $x^2+x-1$ 
 A: The squares in $\mathbb{F}_7$ are indeed
$$
0^2=0,\quad
1^2=1,\quad
2^2=4,\quad
3^2=2,\quad
4^2=2,\quad
5^2=4,\quad
6^2=1,
$$
so you get $\{0,1,2,4\}$.
The cubes are
$$
0^3=0,\quad
1^3=1,\quad
2^3=1,\quad
3^3=6,\quad
4^3=1,\quad
5^3=6,\quad
6^3=6,
$$
so you get $\{0,1,6\}$.
In order to find a field with $7^2$ elements, you just need to find an irreducible degree two polynomial in $\mathbb{F}_7[x]$. For instance $x^2-3$.
The field $\mathbb{L}=\mathbb{F}_7(\sqrt{3})$, where we add a root of $x^2-3$ is, up to isomorphism, the unique field with $49$ elements. In particular, also $x^2+x-1$ must have roots in it. A root must be of the form $a+b\sqrt{3}$:
$$
(a+b\sqrt{3})^2+(a+b\sqrt{3})-1=0
$$
becomes
$$
a^2+3b^2+a-1=0,\qquad 2ab+b=0
$$
so either $b=0$ or $a=-1/2=3$. Since $x^2+x-1=0$ has no roots in $\mathbb{F}_7$, the condition $b=0$ must be discarded and we get
$$
3b^2=-a^2-a+1=-9-3+1=3
$$
so $b=\pm1$. Thus the roots are $3+\sqrt{3}$ and $3-\sqrt{3}$.
You get a field with $343$ elements by adding the roots of a cubic irreducible polynomial, for instance $x^3-2$ (it has no roots, so it's irreducible because it has degree $3$).
There is no root of $x^2-3$ in this field, because this field has no subfield with $49$ elements: a field with $7^m$ elements has a subfield with $7^n$ elements if and only if $n\mid m$.
