Definition of an algebra over a noncommutative ring - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:27:53Z http://mathoverflow.net/feeds/question/21899 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring Definition of an algebra over a noncommutative ring Rasmus 2010-04-19T21:54:26Z 2010-04-21T17:59:25Z <p>I've tried in vain to find a definition of an algebra over a <em>noncommutative</em> ring. Does this algebraic structure not exist? In particular, does the following definition from <a href="http://en.wikipedia.org/wiki/Algebra_(ring_theory)" rel="nofollow">http://en.wikipedia.org/wiki/Algebra_(ring_theory)</a> make sense for noncommutative $R$?</p> <blockquote> <p>Let $R$ be a commutative ring. An algebra is an $R$-module $A$ together with a binary operation $$[\cdot,\cdot]: A\times A\to A$$ called $A$-multiplication, which satisfies the following axiom: $$[a x + b y, z] = a [x, z] + b [y, z], \quad [z, a x + b y] = a[z, x] + b [z, y]$$ for all scalars $a$, $b$ in $R$ and all elements $x$, $y$, $z$ in $A$.</p> </blockquote> <p>So, is there a common notion of an algebra over a noncommutative ring?</p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring/21900#21900 Answer by captain obvious for Definition of an algebra over a noncommutative ring captain obvious 2010-04-19T22:05:09Z 2010-04-19T22:05:09Z <p>Why not just: $A$ is a ring together with a ring homomorphism $R\to A$?</p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring/21905#21905 Answer by mathphysicist for Definition of an algebra over a noncommutative ring mathphysicist 2010-04-19T22:50:23Z 2010-04-19T22:50:23Z <p>Unfortunately that's not <em>exactly</em> what you want but in <a href="http://dx.doi.org/10.1016/j.aim.2008.03.003" rel="nofollow">this paper</a> the authors define <strong>Lie</strong> algebras over noncommutative rings.</p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring/21912#21912 Answer by Manny Reyes for Definition of an algebra over a noncommutative ring Manny Reyes 2010-04-19T23:23:18Z 2010-04-19T23:38:54Z <p>[<strong>Edit</strong>: Thanks to Harry Gindi for pointing out that I'm only extending the notion of an <em>associative</em> algebra.]</p> <p>Here is a somewhat "brute force" approach to define an <em>associative</em> algebra over a general ring. Let's say that a ring $A$ with a fixed ring homomorphism $f\colon R\to A$ is <em>centrally generated</em> over $R$ (with respect to $f$) if $A$ is generated as a ring by the image of $R$ and a subset $X\subseteq A$ such that every element of the image of $R$ commutes with every element of $X$. </p> <p>Then it's clear that whenever $R$ is commutative, a ring homomorphism $f\colon R\to A$ makes $A$ into an $R$-algebra if and only if $A$ is centrally generated over $R$ with respect to $f$.</p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring/21913#21913 Answer by Ben Webster for Definition of an algebra over a noncommutative ring Ben Webster 2010-04-20T00:26:06Z 2010-04-20T00:26:06Z <p>I would argue that this notion doesn't have one natural generalization. One obvious one is what I would call an $R$-bimodule algebra, that is, an algebra which is an R-bimodule in such a way that left multiplication of $A$ commutes with right multiplication of $R$ and vice versa. If you only have one of these actions, you would have an R-left or right module algebra.</p> <p>On some level, you can't really expect there to be one correct generalization; which one is right depends on the context. If you have an example, pick the definition that fits your example, and if you don't have an example, why worry? </p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring/21927#21927 Answer by Zoran Škoda for Definition of an algebra over a noncommutative ring Zoran Škoda 2010-04-20T03:51:08Z 2010-04-21T17:59:25Z <p>The commutative notion of an (associative or not) algebra $A$ over a commutative ring $R$ has two natural generalization to the noncommutative setup, but the one you list with defined <strong>left</strong> $R$-linearity in both arguments is neither of them; in particular your multiplication does not necessarily induce a map from the tensor product, unless the image of $R$ is in the center. Most useful is the notion of an $R$-ring $A$ (or a ring $A$ over $R$), which is just a monoid in the monoidal category of $R$-bimodules: in other words the multiplication is a map $A\otimes A\to A$ which is <em>left</em> linear in first and <em>right</em> linear in the second factor. If we drop the associativity for the multiplication all works the same way, but I do not know if there is a common name (maybe descriptive like magma internal to the monoidal category of $R$-bimodules; or one may try a rare term nonassociative $R$-ring). </p> <p>In the commutative case, the mutliplication is both left and right linear in each factor, what is here possible only if $R$ maps into the center of $A$. (Edit: I erased here one additional nonsense sentence clearly written when tired ;) ). Thus the two useful concepts in the noncommutative case are $R$-rings (possibly nonassociative!) and, well, the subclass with that property: $R$ maps into $Z(A)$, deserving the full name of "algebra". There is also a notion of $R$-<a href="http://ncatlab.org/nlab/show/coring" rel="nofollow">coring</a>, which is a comonoid in the monoidal category of $R$-bimodules, generalizing the notion of an $R$-coalgebra to a noncommutative ground ring.</p> <p>Edit: I suggest also this <a href="http://golem.ph.utexas.edu/category/2008/12/a_quick_algebra_quiz.html" rel="nofollow">link</a>.</p>