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I have an elliptic curve $E$ defined over a field $\mathbb{F}_p$ for $p$ prime and I need to let $\overline{\mathbb{F}_p}$ be the closure of $\mathbb{F}_p$ and then choose points $P\in E\left(\overline{\mathbb{F}_p}\right)$

In this question I was told that this is equivalent to:

choosing points in finite algebraic extensions of your $\mathbb{F}_p$ (since the algebraic closure is just the union of these) so what you really want is to construct these finite extensions and work with them

however my question now is how do I go about constructing the finite extension of a field $\mathbb{F}_p$?

I am self-teaching so assume limited knowledge of complex maths and theorems

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All finite extension of $\mathbb{F}_p$ are of the form $$ \mathbb{F}_p[\alpha], $$ where $\alpha$ is a root of a monic, irreducible polynomial $f(x) \in \mathbb{F}_p[x]$ of some degree $n$.

Such an extension has order $p^{n}$. Given $p$ and $n$, there is up to isomorphism exactly one such extension.

Calculating in such an extension goes like this. Elements can be uniquely written in the form $$\tag{elts} a_{0} + a_{1} \alpha + \dots + a_{n-1} \alpha^{n-1}, $$ for $a_{i} \in \mathbb{F}_p$. Sums are straighforward.

As to products, you work them out and then use the fact that $\alpha$ is a root of $f$ to reduce to the form (elts). In other words, elements (elts) are of the form $g(\alpha)$, where $g(x) \in \mathbb{F}_p[x]$ has degree less than $n$ (including $g(x) = 0$). To calculate $g(\alpha) h(\alpha)$, you first compute the product $g(x) h(x)$ of the two polynomials, divide by $f(x)$ to get $$ g(x) h(x) = q(x) f(x) + r(x) $$ for some $q(x), r(x)$, where $r(x)$ has degree less than $n$ (or $r(x) = 0$), and then $g(\alpha) h(\alpha) = r(\alpha)$ will be in the form (elts).

This is a $\text{crash}^{n}$ course in finite fields. The Wikipedia has much more, including worked out examples.

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