Let $X$ be a variety over perfect field $k$ and $x \in X$ some closed reduced point. (at this point I'm not 100% percent sure if it's neccessary to assume $x$ to be reduced, ie that it's stalk is reduced)
My question is following:
Is there a relatively 'elementary' argument to show the existence of a regular hypersurface $Z \subset X$ containing $x$?
What I tried:
Idea no 1: Since $k$ is perfect field a closed point of $X$ is regular if and only if it is smooth. The locus of regular points is not empty and the locus of smooth points is open, therefore under the locus of regular points $X_{reg} \subset X$ is open too. Inside $X_{reg}$ we can always find a regular hypersurface $Z' \subset X_{reg} $. Even the approach appears to me to be most natural and elementary there there is an obvious problem I not know how to overcome: Does this $Z'$ extends to regular hypersurface $Z \subset X$? If $x$ is not regular in $X$ this approach may not work.
Idea no 2: If $X$ is quasi-projective we can pass to algebraic closure
$X_{\overline{k}}:= X \times \text{Spec} \ \overline{k}$
and invoke the Theorem of Bertini which says that for
quasi-projective $X_{\overline{k}} \subset \mathbb{P}^N_{\overline{k}}$
a general hyperplane section $H \cap X_{\overline{k}}$ is smooth. Any
hyperplane $H \subset \mathbb{P}^N_{\overline{k}}$ defined over $k$
with $H \cap X_{\overline{k}}$ smooth solves the problem.
Doubts if that's the most quick approach : I'm not sure if Theorem of Bertini
gives the most simple way
to find such a regular hypersurface $Z \subset X$ containing $x$, because
it gives an much stronger statement that $Z$ can be choosen to be
a hyperplane section $H \subset X$ with a $H \subset \mathbb{P}^N$.
But hot every hypersurface comes as hyperplane section, so it may suggest thatthere may be a more 'elementary' way to construct such
regular hypersurface $Z \subset X$ containing $x$ without invoking too 'deep' facts.
Motivation: In a comment here Qing Liu wrote that it not always possible to find such hypersurface if the field $k$ is imperfect, and in turn the natural question is why such regular hypersurface $Z \subset X$ containing $x$ always exist if $k$ is perfect? And how 'obvious' is it to see this?
