Let $k$ be an algebraically closed field of characteristic $p>0$, $A$ a regular local noetherian $k$-algebra, $B$ another local noetherian $k$-algebra, $f:A \to B$ an injective ring homomorphism of noetherian local rings, $K$ the fraction field of $A$ and $\bar{K}$ an algebraic closure of $K$. Suppose that the tensor product, $\bar{K} \otimes\_A B$ contains nilpotent elements.
Does there exist a finite field extension $L$ of $K$ such that $L \otimes\_A B$ contains nilpotent elements and $\bar{K} \otimes\_A ((L \otimes\_A B)/\mathrm{Nil})$ contains no nilpotent element, where $\mathrm{Nil}$ is the nilradical of $L \otimes\_A B$?


Is there any reference where I can read on this topic?



Field extension and nilpotent element
This is false in general, even when $A$ and $B$ are fields.


**Example.** Let $A = K = k(t)$, and $B = k(t^{\frac{1}{p^\infty}}) \cong A[\{x\_i\}\_{i \geq 0}]/(x\_0 - t, \{x\_{i+1}^p - x\_i\}\_{i \geq 0})$. Thus,
\begin{align\*}
B \otimes\_A \bar K &= \bar K[\{x\_i\}\_{i \geq 0}]/(x\_0 - t, \{x\_{i+1}^p - x\_i\}\_{i \geq 0})\\
&\cong \bar K[\{z\_i\}\_{i \geq 0}]/(z\_0, \{z\_{i+1}^p-z\_i\}\_{i \geq 0}),
\end{align\*}
through the identification $z\_i = x\_i - t^{\frac{1}{p^i}}$. This has infinitely many nilpotents $z\_i$ (with $z\_i^{p^i} = 0$), and clearly there is no single $L$ over which the $z\_i$ are defined.


**Remark.** On the other hand, the answer is positive if $f$ is essentially of finite type (i.e. $B$ is a localisation of a finite type $A$-algebra). Indeed, in this case $B \otimes\_A \bar K$ is essentially of finite type over $\bar K$, hence Noetherian. Thus, the nilradical $\mathfrak{rad}(B \otimes\_A \bar K)$ is finitely generated; say by $x\_1,\ldots,x\_r$.


Let $L$ be a field over which all the generators are defined, i.e. there exist $y\_1,\ldots,y\_r \in B \otimes\_A L$ such that the image of $y\_i$ in $B \otimes\_A \bar K$ is $x\_i$. Note that each $y\_i$ is nilpotent, for example since $B \otimes\_A L \to B \otimes\_A \bar K$ is injective and each $x\_i$ is nilpotent.


Now consider $C = (B \otimes\_A L)/(y\_1,\ldots,y\_r)$. I claim that $C$ is geometrically reduced (as $L$-algebra) (see [Tag 05DS](http://stacks.math.columbia.edu/tag/05DS) for this notion). Indeed, when we apply $- \otimes\_L \bar K$, we get $(B \otimes\_A \bar K)/(x\_1,\ldots,x\_r)$, which is reduced by our choice of the $x\_i$. $\square$


(As a corollary, we get that $C$ is reduced, so $y\_1,\ldots,y\_r$ are the only nilpotents in $B \otimes\_A L$.)


**Remark.** Note that I never use that $A$ and $B$ are local, Noetherian, regular, etc. It is rather a relative property of the morphism $f$ (in fact, I only need the morphism $f\_K \colon K \to B \otimes\_A K$).


