# Alternativity on $A \otimes B$

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?

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

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.

• 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...
– Dac0
Nov 16, 2021 at 21:23
• @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. Nov 17, 2021 at 8:08
• 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?
– Dac0
Nov 17, 2021 at 9:43
• @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). Nov 17, 2021 at 10:12
• yes, you are right, that's why doesn't work that way. Thank you :)
– Dac0
Nov 17, 2021 at 11:09