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Let $k$ be a field of characteristic $p > 0$. Let $K$ be an extension field of $k$. Let $k^{p^{-\infty}}$ be the perfect closure of $k$. Then Spec($K\otimes_k k^{p^{-\infty}}$) is a one element set?

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    I could not log in MathOverflow at that time so I could not upvote for Robin Chapman. I have no idea why my comment saying so was removed by a moderator. By the way, I already upvoted for him.2013-10-22

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Let $A = K\otimes_k k^{p^{-\infty}}$. We identify $K$ with its canonical image in $A$. Let $P$ and $Q$ be prime ideals of $A$. It suffices to prove that $P \subset Q$. Let $\pi\colon A \rightarrow A/P$ be the canonical homomorphism. Let $\sigma\colon K \rightarrow A/P$ be the restriction of $\pi$. Let $x \in P$. It suffices to prove that $x \in Q$. $x$ can be written as $x = \sum_{i} x_i \otimes y_i$, where $x_i \in K$ and $y_i \in k^{p^{-\infty}}$. There exists an integer $n \gt 0$ such that $y_i^{p^n} \in k$ for all $i$. Then $x^{p^n} = \sum_{i} x_i^{p^n} \otimes y_i^{p^n} = \sum_{i} x_i^{p^n}y_i^{p^n} \otimes 1$. Since $x^{p^n} \in P$, $\pi(x^{p^n}) = \sum_{i} \sigma(\sum_{i} x_i^{p^n}y_i^{p^n}) = 0$. Hence $\sum_{i} x_i^{p^n}y_i^{p^n} = 0$. Hence $x^{p^n} = 0$. Hence $x^{p^n} \in Q$. Since $Q$ is prime, $x \in Q$ as desired. QED