what is the definition of the Picard group of a (non necessarilly commutative) Ring? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T22:51:50Z http://mathoverflow.net/feeds/question/109957 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109957/what-is-the-definition-of-the-picard-group-of-a-non-necessarilly-commutative-ri what is the definition of the Picard group of a (non necessarilly commutative) Ring? Hector Pinedo 2012-10-17T21:51:33Z 2012-10-18T06:31:00Z <p>Hi. I have only able to find the definition of $Pic(R)$ for a commutative ring $R.$ Which is the isomorphism classes of projective $R$-modules of rank $1,$ and the product given by $[A][B]=[A\otimes_R B].$</p> <p>How can i adapt this definition for the non-commutative case?</p> <p>are there some books of articles related to this ? Thanks.</p> http://mathoverflow.net/questions/109957/what-is-the-definition-of-the-picard-group-of-a-non-necessarilly-commutative-ri/109966#109966 Answer by arsmath for what is the definition of the Picard group of a (non necessarilly commutative) Ring? arsmath 2012-10-18T01:54:27Z 2012-10-18T01:54:27Z <p>Amnon Yekutieli has studied the derived Picard group in a noncommutative setting. See <a href="http://arxiv.org/abs/math/9810134" rel="nofollow">Dualizing Complexes, Morita Equivalence and the Derived Picard Group of a Ring</a>.</p> http://mathoverflow.net/questions/109957/what-is-the-definition-of-the-picard-group-of-a-non-necessarilly-commutative-ri/109980#109980 Answer by Stefan Waldmann for what is the definition of the Picard group of a (non necessarilly commutative) Ring? Stefan Waldmann 2012-10-18T06:31:00Z 2012-10-18T06:31:00Z <p>The definition I appreciate most comes from Morita theory: you consider unital (just for simplicity, something slightly weaker will also work) rings and the bimodules between them. Using the tensor product this makes almost a category: the tensor product of bimodules (over the ring in the middle) is associative up to a natural isomorphism and the rings themselves, viewed as bimodules over themselves with the mutliplication being the left/right module structures constitute the units, again up to a natural isomorphism. So either you prefer a bicategory setting, then you are done with what I said, or you prefer an honest category, then you have to pass to isomorphism classes of bimodules as morphisms between the rings. Of course, there are the usual set-theoretic issues that you should work in some universe etc.</p> <p>But now the definition of the Picard groupoid is very simple: it is the groupoid of invertible arrows in this category. The Picard group of a single ring is then just the isotropy group at this ring of the big Picard group. In other words: the Picard group of $R$ is the group of invertible bimodules with respect to the tensor product as multiplication and the ring $R$ as unit.</p> <p>It is then a famous theorem of Morita that characterizes the invertible bimodules $M$ between $R$ and $S$: there are various versions but a simple one is that they are finitely generated projective right modules over $R$ such that if you write $M = eR^n$ with an idempotent $e \in M_n(R)$ one has $ReR = R$, i.e. the two-sided ideal generated by the entries of the matrix $e$ is the whole ring $R$. Then the ring $S$ is given (via the left module structure) by the $R$-linear endomorphisms of $M$.</p> <p>You can find this in many algebra textbooks like e.g. Lam's book on Modules and Rings, or Bass' book on Algebraic $K$-Theory. On my homepage you can also find some (very preliminary) lecture notes covering this stuff and other versions of Morita theory for rings with additional structures.</p>