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A ring is semiprime if it has no non-zero 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.

Michael Barr
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  • This is (1) in http://mathoverflow.net/questions/9436/reduced-%E2%8A%97-reduced-reduced-what-about-connected , right? (I am still looking for a constructive proof, by the way. The similar-sounding question http://mathoverflow.net/questions/250040/constructive-proof-that-a-kernel-consists-of-nilpotent-elements?rq=1 has been nicely resolved using topos theory. Maybe you can help constructivizing the non-constructive proof from Milne's AG?) – darij grinberg Dec 14 '16 at 17:55
  • 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). – R. van Dobben de Bruyn Dec 15 '16 at 03:41

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