Let $X$ be an irreducible locally noetherian $k$-scheme ($k$ any field) and $G$ an algebraic $k$-group acting on $X$. Proposition 3.1.6 in these notes by M. Brion claims
Let $a : G \times X \to X$ be an action of an algebraic group on an irreducible locally noetherian scheme and let $x \in X(k)$. Then the quotient group scheme $C_G(x)/\text{Ker}(a)$ (quotient of centralizer) is affine.
The proof easily reduces to the case $G=C_G(x)$ and hence that $x \in X(k)$ is fixed by $G$. Let's consider in the following also only this case and so specialize the previous claim to:
Let $a : G \times X \to X$ be an action as before and let $x \in X(k)$ be fixed by $G$. Then $G$ is affine.
which Brion after the remarked reduction step actually proves.
Rmk.: This is actually also the statement of Prop. 8.9 in Milne's Algebraic Groups
The both proofs mainly use the observation that as $G$ assumed to fix $x$, the $G$-action induces to an action on the stalk $\mathcal{O}_{X,x}$ and consequently on infinitesimal neighborhoods $x_{(n)}:=\operatorname{Spec}(\mathcal{O}_{X,x}/ \mathfrak{m}_x^n)$ and so forth deriving at the end that certain fixed subscheme $X^H \subset X$ must be already whole $X$, giving at the end that $G$ embeds as closed immersion in appropr $\text{GL}_m$.
Question: Why it is neccessary for the given proofs to assume the closed fixed point $x$ by $G$ to be rational, ie $x \in X(k)$, and so in turn why it cannot be for example a closed $L$- valued point $x \in X(L)$ for some finite field extension $k \subset L$?
(EDIT #1: Note that $L$ has to be finite (equivalently by Zariski's lemma that $x$ is closed in $X$ in topol. sense) is nessessary as the proof proceeds by constructing finite dimnl reps over $k$ of local rings $\mathcal{O}_{X,x}/ \mathfrak{m}_x^n$. So as @Dave Benson pointed out that eg $x$ choosen to be the generic point runs the proof mechanism there badly into troubles as the proof's strategy there includes producing only finite dimensional $k$-reps.
In other words, where precisely the argumentation of the presented proofs would break down if $x$ would not be rational?
If I'm not missing something going through the presented proof, then assuming this the $G$ action would still descend to the induced action on infinitelimal neighborhoods of $x$, and also all other steps in the proof seemingly nowhere exploit to rationality $x$, so seemingly the proof would still go through for a nonrational fixed point, or not?
So I not understand where we really need unavoidably that the fixed point $x$ was assumed to be rational.
EDIT #2: As LSpice pointed out to prove the claim as it is stated there it indeed suffice to assume that the fixed point is rational applying fppf descent (...affiness can be checked on fppf cover), so by base changing if neccessarily it suffice to treat the case that the fixed point is $k$-rational.
But as I emphsised in the comments my question is primary about the "legitimation" of the "methods/techniques" of the given proof depending on $x$ is rational or not. Essentially, the leading question should be phrased as that what if the fixed closed point $x$ is not rational, but we still carry out "literally with head trought the wall" the same argumentation as Brion resp. Milne gave in the notes except that $x$ is not rational? Then, which steps precisely in the proof mechanism would fail or even cannot be conducted or even phrased?
Maybe following LSpice's comments the problem is also that certain constructions used there have no meaning when $x$ is not rational. But I not sure if that's the case, probably I misunderstood LSpice's point below in the comments.
Let me try to elaborate it in more details, maybe this reveals also what not understood. If closed $x$ is fixed by $G$, then formally by "the action on the local ring $\mathcal{O}_{X,x}/ \mathfrak{m}_x^n$ should be formulated in terms of point functors namely as map $G \to \operatorname{Aut}_k(\mathcal{O}_{X,x}/ \mathfrak{m}_x^n)$ defined for every fin. gen. $k$-algebra $R$ by group homomorphism $G(R) \to \operatorname{Aut}(R \otimes_k (\mathcal{O}_{X,x}/ \mathfrak{m}_x^n)$.
Intermediate question: Is it at the stage neccessary that $x$ is rational in order that this action is well defined? I not see here any problem but maybe that's exactly the problematic step.
So the question is if there is something "fishy" with this functor if $x$ would be not rational?
EDIT #3: above appears also a potential notation problem/ notation abusion with stalk $\mathcal{O}_{X,x}$ for $x \in X(L)$ non rational: keeping the notation precise by the stalk $\mathcal{O}_{X,x}$ is actually meant the stalk $\mathcal{O}_{X,\mathfrak{m}_x}$ which is as usually formed by localizing appropr ring $R$ corresponding to an affine chart of $X$ containing image of $x$ at (maximal) prime $\mathfrak{m}_x$ be the kernel of $x^*: R \to L$ corresponding to $x$.
