## Non-trivial integral forms of algebras

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.

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?

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 Minor quibble: I think you want $\mathcal B$ to be something like free as $\mathbb Z$-module. Otherwise $\mathcal B$ could be for instance a $\mathbb Q$-vector space (note that for a $\mathbb Q$-algebra $\mathcal B$ we have $\mathcal B\bigotimes_{\mathbb Z}\mathbb C=\mathcal B\bigotimes_{\mathbb Q}\mathbb C$. – Torsten Ekedahl Jul 24 2011 at 4:51

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 $g$.

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!