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0 pointsjoppy5y ago0 comments

I think that usually when people say “algebra over a ring” they assume that ring to be commutative, so that the word “bilinear” in “bilinear multiplication” is useful. It’s possible to define an algebra over a non-commutative ring as a bimodule (rather than left module or right module) equipped with a bilinear multiplication, but I have rarely seen this used.

The definition the parent poster used (or intended to use, but wrote the wrong way around, I believe) was that an algebra over a non-commutative ring is just an algebra over its commutative centre. (In which case, we’re still really just talking about algebras over commutative rings).

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JadeNB5y ago

But the definition doesn't work even for commutative rings; as I mention, it says that the only ℤ-module structure on ℤ/2ℤ is the trivial one, which is not the usual understanding of the term. I agree that, if you switch A and R in Hom(A, Z(R)), then an element of the Hom space Hom_{ring}(R, Z(A)) makes A into an R-algebra, but I would argue it's not the only way; there's a map Hom_{ring}(R, Z(A)) -> Hom_{ring}(R, End_{gp}(A)), but it need not be surjective if the rings aren't assumed unital. Consider, for example, a polynomial ring R = k[t] and its ideal A = tR, which has a natural structure of an R-algebra.

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0 pointsjoppy5y ago0 comments

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JadeNB5y ago

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