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Theorem $9.120$ from Rotman's homological algebra states that if $A, B$ are central simple $k$-algebra, then so is $A \otimes _k B$.

In the proof of $9.133$ it is stated that if $A$ is an central simple $k$-algebra and $E/k$ is a field extension then by Theorem $9.120$ that $E \otimes_k A $ is central simple $E$-algebra.

So here we view $A$ as a central simple $E$-algebra to use the theorem? Can $A$ even be viewed this way? Any help would be appreciated!

scsnm
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  • No, it's $E\otimes_k A$ that's the CSA over $E$, not $A$. – Angina Seng Aug 11 '20 at 15:14
  • I am not sure how Rotman used 9.120 as $A$ is only a $k$-algebra here. If $A$ is not a CSA over $E$, then 9.120 doesn't apply, right? – scsnm Aug 11 '20 at 15:16
  • I don't see what the problem is.... – Angina Seng Aug 11 '20 at 15:20
  • Maybe I am being silly here. 9.120 says for the tensor to be CSA over $k$ both A and B have to be CSA over $k$. But in 9.133 A is not CSA over $E$, only E is. so 9.120 still applies? – scsnm Aug 11 '20 at 15:22
  • Here is what's true. If $A$ and $B$ are finite dimensional simple $k$-algebras with centre $k$ then so is $A\otimes_k B$. Also $E\otimes_k A$ is a finite-dimensional simple $E$-algebra with centre $E$. – Angina Seng Aug 11 '20 at 15:34
  • I see. Thank you Angina! – scsnm Aug 11 '20 at 15:38

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