In the paper "Algebraic Attacks on the Courtois Toy Cipher" written by M. Albrecht, he defined field polynomials and stated a corollary as follows:
Definition: Let $k$ be a field with order $q=p^n$, $p$ prime and $n>0$. Then the field polynomials of the ring $k[x_0,...,x_{n-1}]$ are defined as the set $$\{x_0^q-x_0,\dots, x_{n-1}^q-x_{n-1}\}$$ The ideal spanned by this set $$\langle x_0^q-x_0,\dots, x_{n-1}^q-x_{n-1}\rangle$$ is called the field ideal of $k[x_0,\dots,x_{n-1}]$.
Corollary: Let $I$ be an ideal in $k[x_0,\dots,x_{n-1}]$. The ideal spanned by the generators of $I$ and generators of the field ideal has the same variety over $k$ as the ideal $I$ but excludes all elements from $\bar{k}$, the algebraic closure of $k$.
The paper provided the proof that the finite field $k$ satisfies $x^q=x$ for every $x$. Thus the equations $x_i^q-x_i=0$ are satisfied for every possible value in $k$ and especially for the elements in $V(I)$. Also it factors completely over $k$ and thus no elements of $\bar{k}$ satisfies it.
My question: I understand the first part of the proof and the argument of no elements of $\bar{k}$ satisfies it.
I am not sure why the appending of the field ideal would remove anything from the original $V(I)$. Maybe I am still confused about the concept of algebraic closure. Can anyone give a concrete example where $V(I)$ contains something from $\bar{k}$, which does not satisfy the field polynomials?
Any help is appreciated!
You'll see this kind of thing creep into a lot of contexts where people try to distill their ideas into a more elementary setting. It's a cousin to situations where people talk about non-reduced structure on a complex manifold, a notion that (as stated) is gibberish.
In particular, if you think about things scheme-theoretically, the variety corresponding to an ideal $I < k[x_0, \ldots, x_{n-1}]$ has points whose coordinates belong to $F$, but it also has points whose coordinates belong to larger fields. This is something that automatically falls out of the definition of a scheme. (I'll avoid making this precise, because it depends on whether you're talking about "set-theoretic points" or "geometric points" or what.)
Since moving to extension fields is a pretty natural thing to do in number theory, it's common for number theorists and others to talk very vaguely about this sort of thing. To someone experienced with schemes it's fairly easy to translate what's said above into a precise statement, but to someone without that background statements of this form can be nearly incoherent.
As has apparently been mentioned by @user26857 and @Roland in the comments, what's meant is something like this. Given a set of equations with coefficients in $\mathbb{F}_{q}$, we can consider its solutions in $\mathbb{F}_{q}$, but we can also consider its solutions in any extension field of $\mathbb{F}_q$. If we want to consider all such solutions at once, it's enough to consider the solutions in $\overline{\mathbb{F}_q} = \overline{\mathbb{F}_p}$. So whenever we think of a variety over $\mathbb{F}_p$, we automatically and without further mention think of it as somehow being a subset of the corresponding variety over $\overline{\mathbb{F}_p}$.
However, since the elements of $\mathbb{F}_q$ are precisely those elements $s \in \overline{\mathbb{F}_q}$ for which $s^q = s$, if we throw additional equations of the form $x_i^q = x_i$ into our system then there will be no "new' points when we look at solutions over $\overline{\mathbb{F}_p}$ rather than over $\mathbb{F}_q$.