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I have $A$ an associative algebra and $B$ at least an alternative algebra. Is there a sufficient condition on $A$ or $B$ to have $A \otimes B$ an alternative algebra?

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    $\begingroup$ Please do not crosspost. $\endgroup$ Nov 16, 2021 at 19:18
  • $\begingroup$ Yes, sorry. Thank you @DietrichBurde $\endgroup$
    – Dac0
    Nov 16, 2021 at 21:01

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The monograph “Alternative Loop Rings” by Goodaire, Jespers and Polcino Milies (North Holland Mathematics Studies 184, 1996) contains, in chapter I (“Alternative Rings”), §5 (“Tensor Products”), the following proposition (5.13; I'm changing the notation to match yours):

Let $B$ be an alternative algebra over a field $F$ and suppose $A$ is a commutative associative algebra over $F$. Then the tensor product $A\otimes_F B$ is alternative.

So $A$ being commutative is a sufficient condition.

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  • $\begingroup$ The question was correctly answered. I then ask you if you are aware if this condition on A is necessary or if there are other conditions on A or B that are sufficient. Unfortunately “Alternative Loop Rings” doesn't seem to address the question in full generality... $\endgroup$
    – Dac0
    Nov 16, 2021 at 21:23
  • $\begingroup$ @Dac0: Well, $B$ being associative is evidently sufficient, and this shows that $A$ being commutative is not necessary. But I really don't know anything beyond this, and I suspect not much is known. $\endgroup$
    – Gro-Tsen
    Nov 17, 2021 at 8:08
  • $\begingroup$ Thank you @Gro-Tsen what is humbling for me is that if $B$ being associative is sufficient for $A \otimes B$ being alternative, then when I have two elements $a \otimes b$ and $c \otimes d$ on $A \otimes B$ I could restrict myself to the associative subalgebra of $B'$ cointaining $c$ and $d$ that is associative and therefore the subalgebra generated by $a \otimes b$ and $c \otimes d$ would be associative itself. Then we would have the algebra $A \otimes B$ being alternative every time $A$ associative and $B$ alternative. Am I missing something? $\endgroup$
    – Dac0
    Nov 17, 2021 at 9:43
  • $\begingroup$ @Dac0: What you are missing is simply that elements of $A\otimes B$ are not necessarily of the form $a\otimes b$ (i.e., pure tensors). $\endgroup$
    – Gro-Tsen
    Nov 17, 2021 at 10:12
  • $\begingroup$ yes, you are right, that's why doesn't work that way. Thank you :) $\endgroup$
    – Dac0
    Nov 17, 2021 at 11:09

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