Non-trivial integral forms of algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:55:55Z http://mathoverflow.net/feeds/question/61743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61743/non-trivial-integral-forms-of-algebras Non-trivial integral forms of algebras Najdorf 2011-04-14T20:34:03Z 2011-07-24T01:59:43Z <p>Suppose $\mathcal{A}$ is a $\mathbf{C}$-algebra then an integral form would be a subring $\mathcal{B} \subset \mathcal{A}$ such that the canonical map $\mathcal{B} \otimes_{\mathbf{Z}} \mathbf{C} \rightarrow \mathcal{A}$ is a bijection. </p> <p>For some algebras there is an obvious integral form in the following sense: there is a preferred $\mathbf{C}$-basis for $\mathcal{A}$ and the $\mathbf{Z}$-span of that basis is $\mathcal{B}$. Now my question is do we have examples where $\mathcal{B}$ is non-obvious? In other words the basis coming from $\mathcal{B}$ would look very strange for those who only work with $\mathcal{A}$. Is there such an example where $\mathcal{A}$ is commutative?</p> http://mathoverflow.net/questions/61743/non-trivial-integral-forms-of-algebras/71093#71093 Answer by Chris for Non-trivial integral forms of algebras Chris 2011-07-24T01:54:11Z 2011-07-24T01:54:11Z <p>There are many important integral forms with a "strange" appearance. Namely, we have 1) Kostant's form for a finite-dimensional complex simple Lie algebra, see the classical Humphrey's book; 2) Garland's form for the loop algebra of a finite-dimensional complex simple Lie algebra $g$, see the proper Garland's paper. In this case appear some elements given by the coefficient of some series given by the exponential of a suitable generating series with coeficients in the loop Cartan subalgebra of $g$.</p> <p>For commutative cases we probably have those expected elements from a $\mathbb{C}$-basis, since the "strange" elements always come from brackets in the non-commutative case!</p> <p>Hope help you,</p> http://mathoverflow.net/questions/61743/non-trivial-integral-forms-of-algebras/71094#71094 Answer by Mariano Suárez-Alvarez for Non-trivial integral forms of algebras Mariano Suárez-Alvarez 2011-07-24T01:59:43Z 2011-07-24T01:59:43Z <p>Finite dimensional algebras (over alg. closed fields) which are of finite representation type have <em>multiplicative bases</em>, that is, bases such that the product of two of its elements is either an element of the basis or zero. This is a difficult and beautiful theorem of R. Bautista, P. Gabriel, A. Roiter and L. Salmerón.</p> <p>This gives rather special integral forms for such algebras and, in general, it is quite non-obvious how to get them.</p>