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?


1 Answer 1


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 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 $f_K \colon K \to B \otimes_A K$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.