Image points over extensions fields While trying to understand isogeny-based cryptography I stumbled over a passage in a paper by Galbraith. On p.4 he explains how to calculate the image points of an isogeny $\phi:E\to E^{\prime}$:

What about the other 4 points in $E^{\prime}(\mathbb{F}_7)$, such as (3, 0)? These are the image of points on $E$ over an extension of $\mathbb{F}_7$. Consider the point $Q = (1, \alpha)\in E(\mathbb{F}_{7^2})$ where $\alpha \in \mathbb{F}_{7^2}$ satisfies $\alpha^2 = 3$. One can check that $Q$ has order 4,
$[2]Q = (0, 0)$, and $\phi(Q) = (3, 0)$.

I'm not sure what $\mathbb{F}_{7^2}$ looks like. If it is a Galois field over a polynomial, how would you translate the polynomials back into $x,y$ values for a point in $\mathbb{F}_{7}$?
 A: $\mathbb{F}_{7^2}$ is a vector space over $\mathbb{F}_7$ of dimension $2$, so you just need to find a basis.
There are plenty of representations of $\mathbb{F}_7$ (as many as the number of irreducible polynomials of degree $2$ over $\mathbb{F}_7$), but Galbraith uses $\mathbb{F}_{7^2}\simeq\mathbb{F}_7[X]/(X^2-3)$. This isomorphism holds because $X^2-3$ is irreducible over $\mathbb{F}_7$ (it has degree $2$ and $3$ is not a square mod $7$).
Hence, an $\mathbb{F}_7$ basis is given by $1,\alpha$, where $\alpha$ satisfies $\alpha^2=3$ ($\alpha=$ the class of $X$, or if you prefer , any root of $X^2-3$ in a fixed algebraic closure of $\mathbb{F}_7$.)
All in all, any element of $\mathbb{F}_{7^2}$ may be written in a unique way as $a+b\alpha,a,b\in\mathbb{F}_7$.
Addition and multiplication are carried on as you may think. 
To be precise: $(a+b\alpha)+(c+d\alpha)=(a+c)+(b+d)\alpha$, and 
$(a+b\alpha)(c+d\alpha)=ac+bd\alpha^2+bc\alpha+ad\alpha=(ac-3bd)+(ad+bc)\alpha$.
Moreover, an element $a+b\alpha$ of $\mathbb{F}_{7^2}$ will lie in $\mathbb{F}_7$ if and only if $b=0$. 
