A ring is semiprime if it has no nonzero nilpotents. Let $Q$ denote the rational numbers and $A,B$ be a pair of commutative semiprime $Q$ algebras. Is $A\otimes_Q B$ semiprime? It is well known that the tensor product of two fields in finite characteristic can have nilpotents.
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$\begingroup$ This is (1) in mathoverflow.net/questions/9436/… , right? (I am still looking for a constructive proof, by the way. The similarsounding question mathoverflow.net/questions/250040/… has been nicely resolved using topos theory. Maybe you can help constructivizing the nonconstructive proof from Milne's AG?) $\endgroup$– darij grinbergDec 14, 2016 at 17:55

$\begingroup$ What you call semiprime is usually called reduced. The positive answer to your question follows directly from Tag 034N and Tag 030V (3) $\Leftrightarrow$ (5) (since $\mathbb Q$ is perfect). $\endgroup$– R. van Dobben de BruynDec 15, 2016 at 3:41
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