Base change of purely inseparable extension If $K$ is a purely inseparable extension  of $k$ and $A$ is a $k$-algebra, since $A\otimes_kK$ may be not a field, hence I want to ask the following question:
If $\mathfrak{p}$ is a prime of $A\otimes_kK$ and $f^{-1}(\mathfrak{p})$ be the preimage of $\mathfrak{p}$ under the natural map $f:A\rightarrow A\otimes_kK$, then how to prove that $\operatorname{Frac}(A/f^{-1}(\mathfrak{p}))\rightarrow\text{Frac}(A\otimes_kK/\mathfrak{p})$ is purely inseparable?
The motivation is that for a scheme over $k$, I want to prove that the induced residue field map of $X_K\rightarrow X$ at any point is purely inseparable.
Thanks in advance.
 A: For completeness, and to get this question off the unanswered queue, here's a fully written out solution. Let $p=\operatorname{char} k$ and let $\overline{x}\big/\overline{y}\in\text{Frac}([A\otimes_k K]\big/\mathfrak{p})$, where $x,y\in (A\otimes_k K)\setminus\mathfrak{p}$. We want to find $n\in\mathbb{N}$ such that $\left(\overline{x}\big/\overline{y}\right)^n$ lies in the image of $\text{Frac}(A\big/f^{-1}(\mathfrak{p}))$, which consists of all elements of the form $\overline{a_1\otimes 1}\big/\overline{a_2\otimes 1}$ for $a_1\in A$ and $a_2\in A\setminus f^{-1}(\mathfrak{p})$.
Now, $x$ and ${y}$ are sums of simple tensors. Suppose ${x}=\sum_{i=1}^na_i\otimes t_i$. Since $K\supseteq k$ is purely inseparable, for each $i$ there is some $m_i$ such that $t_i^{p^{m_i}}\in k$. Thus, letting $m=\sum_{i=1}^nm_i$, we have
\begin{align}
{x}^{p^m}&=(a_1\otimes t_1+\dots+a_n\otimes t_n)^{p^m} \\
&= (a_1\otimes t_1)^{p^m}+\dots+(a_n\otimes t_n)^{p^m} \\
&=a_1^{p^{m}}\otimes t_1^{p^m}+\dots+a_n^{p^m}\otimes t_n^{p^m},
\end{align}
where the omitted terms vanish because we are working in characteristic $p$. Note that $t_i^{p^m}=\left(t_i^{p^{m_i}}\right)^{p^{m-m_i}}\in k$, so each $a_i^{p^m}\otimes t_i^{p^m}=(a_i^{p^m}\big/t_i^{p^m})\otimes 1\in f(A)$ and hence ${x}^{p^m}\in f(A)$; for convenience say $x^{p^m}=a_1\otimes 1$. Similarly, we can find $m'$ such that $y^{p^{m'}}=a_2\otimes 1\in f(A)$; note in particular that, since $y\notin\mathfrak{p}$ and $\mathfrak{p}$ is prime, we have $a_2^{p^{m}}\notin f^{-1}(\mathfrak{p})$. Hence, letting $n=m+m'$, we have $$\left(\frac{\overline{x}}{\overline{y}}\right)^{p^{n}}=\frac{\overline{x^{p^{m+m'}}}}{\overline{y^{p^{m+m'}}}}=\frac{\overline{a_1^{p^{m'}}\otimes 1}}{\overline{a_2^{p^m}\otimes 1}},$$ and so we are done.
