So a) is simple, calculate the squares and cubes mod 7. However, for b) I know that Z7[x]/(f(x)) if f(x) is irreducible of degree 2 would work, but then how do I solve the equations ?
I assume F7xF7 would also contain 7^2 elements but is it possible to solve the equations in that field ?
How would you go for c) also ? Again F7xF7xF7 has 7^3 elements but can you solve in it ?
Thanks.
I'll give full solutions for part (b) only, and hints for parts (c) and (d). Note that I'll construct the fields as quotients in $\mathbb{F}_7[y]$, since the variable $x$ was used in writing out the polynomials.
Part (b)
Let's let $\mathbb{F}_{49} = \mathbb{F}_7[y]/\langle y^2 - 3\rangle$. Then, the elements of $\mathbb{F}_{49}$ are given by $\mathbb{F}_{49} = \{ay+b| a,b \in \mathbb{F}_7\}$. Multiplication is subject to the rule $y^2 - 3 = 0$, so $y^2 = 3$. Thus, the polynomial $x^2 - 3$ factors in $\mathbb{F}_{49}$ as $(x-y)(x+y)$.
Now, let's turn to $x^2-x-1$. We can re-write this as $(x+3)^2 - 3$, so it has roots $x = 3\pm y$.
Part (c) (Hint) Let $f(y) = y^3 - r$, where $r$ is not a cube $\bmod{7}$, and do the usual construction $\mathbb{F}_{7^3} = \mathbb{F}[y]/\langle f(x)\rangle$
Part (d) (Hint)
If $x^2-3$ has roots $\pm\alpha$ in $\mathbb{F}_{7^3}$, then, $\mathbb{F}_{7^3}[x]/\langle x^2-3\rangle$ will have zero divisors (namely, $(\overline{x+\alpha})(\overline{x-\alpha}) = \overline{x^2-3} = \overline{0}$. Using the Third Isomorphism Theorem, we can turn this into a question about roots of $y^3 - r$ in $\mathbb{F}_7$, which is easier to work with.