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?