Normal curve after base change (p > 0) Suppose that $C$ is a normal projective curve over some base field $k$, possibly of positive characteristic. I am wondering to what extent one can modify the base field $k$ while not braking the regularity of $C$. So in particular, is a separable (or maybe separably generated) extension allowed?
My exact use case would be $k\subset K=K(C)$, so $K$ is the function field of the curve itself. Moreover, we can assume that $k$ is algebraically closed in $K$.
 A: This is true for a separable field extension. In fact, this works more generally than just curves. If $X$ is any normal scheme over $k$ and $K/k$ is a separable field extension, then $X_K$ is normal. I'll sketch this proof with appropriate references to technical results in the Stacks Project. 
Normal is a local condition so we can assume $X$ is an affine scheme $\operatorname{Spec}A$ where $A$ is a normal $k$-algebra.  Since $K/k$ is a separable extension then by Lemma 10.145.10, there exists a smooth $k$-algebra $B$ whose fraction field is $K$. We can base change the morphism $k \to B$ by $A$ to get a morphism $A \to B \otimes_k A$. Since $B$ is smooth and smoothness is preserved by base change (Lemma 10.130.4), $A \to B \otimes_k A$ is smooth. By Lemma 10.149.7, $A \otimes_k B$ is normal since $A$ is normal and normality ascends along smooth morphisms. Finally, $A \otimes_k K$ is a localization of $A \otimes_k B$ since $K$ is a localization of $B$. Thus, $A \otimes_k K$ is the localization of a normal ring so it is normal. 
For an example of how this fails for inseparable extensions, let $k = \mathbb{F}_3(t)$ where $t$ is a transcendental. Consider the plane curve $X/k$ given by $y^2z+x^3−tz^3 = 0$. Then $X$ is regular. However, after base extending to $K = \mathbb{F}_3(\sqrt[3]{t})$, $X$ becomes singular with normalization $\mathbb{P}^1$. This example is remark 16 here.
Edit: For curves normal is the same as regular. However, this isn't true in general so we may ask is regularity preserved under separable base change? The answer is yes and the proof is exactly the same as the one above for normality. It turns out that regularity also ascends along smooth morphisms (Lemma 10.149.8), and everything else in the proof above goes through. 
To give a totally complete answer to your question, we can ask for what $X/k$ does $X$ stay regular after base extending to any field extension $K/k$? We can call this geometrically regular. You know from your previous question that this is true if $X$ is smooth. It turns out that this is an if and only if in any reasonably geometric case. Specifically (Lemma 32.10.6), if $X$ is finite type over $k$, then $X$ is geometrically regular if and only if $X/k$ is smooth. 
