Let $R$ be a noetherian domain and let $\mathcal{O}$ be an $R$-algebra that is finitely generated and projective as an $R$-module. The set of invertible fractional ideals of $\mathcal{O}$ is a group under multiplication. Is this group always abelian?



Noncommutative group of invertible ideals of a ring
No; it is not abelian in general.


Let $R$ be a discrete valuation ring with maximal ideal generated by an element $\pi$, let $k := R / \pi R$ be the residue field of $R$ and let $\mathcal{O}$ be the inverse image in $M\_2(R)$ of the scalar matrices in $M\_2(k)$. Note that $\mathcal{O}$ is free of rank $4$ as an $R$-module, with basis $\{e\_{11} + e\_{22}, \pi e\_{11}, \pi e\_{12}, \pi e\_{21}\}$ say. So $\mathcal{O}$ is, in particular, finitely generated and projective as an $R$-module.


For every $g \in GL\_2(R)$, you have the invertible fractional ideal $\mathcal{O}g$. Note that $\mathcal{O}g \cdot \mathcal{O} h = \mathcal{O} gh$ since $GL\_2(R)$ normalises $\mathcal{O}$ inside $M\_2(R)$. Note also that $\mathcal{O} g = \mathcal{O} h$ if and only if $gh^{-1} \in \mathcal{O}^\times$, which is the preimage in $GL\_2(R)$ of the group of scalar matrices in $GL\_2(k)$. So we obtain an injective group homomorphism from $PGL\_2(k) \cong GL\_2(R) / \mathcal{O}^\times$ into the group $I(\mathcal{O})$ of invertible fractional ideals of $\mathcal{O}$.


In the positive direction, if $K$ is the field of fractions of a Dedekind domain $R$ and if $\mathcal{O}$ is a *maximal order* in some finite dimensional separable $K$-algebra, then $I(\mathcal{O})$ is abelian.


See Theorem 12 and Theorem 13(c) of [this](http://www.ams.org/journals/tran/1973-180-00/S0002-9947-1973-0318204-3/S0002-9947-1973-0318204-3.pdf "This") paper by Frőhlich, which gives a thorough discussion of the Picard group of a noncommutative ring. 


