Clifford algebra of a free bimodule over a noncommutative ring I have an apparent definition (and construction) of the Clifford algebra of a free $R$-$R$-bimodule $M$ with a quadratic form $q: M \rightarrow R$ with noncommutative $R$.  I am not aware of any restrictions, although I limit this to bimodules of finite rank.
Let $R$ be a ring.
Let $M$ be a free $R$-$R$-bimodule of finite rank.
Let $b : M \times M\rightarrow R$ be a bilinear form (not necessarily symmetric): $\forall r,s \in R, x,y\in M: b(r.x,y.s)=r\ b(x,y)\ s$.
Let $q:M\rightarrow R:x\mapsto b(x,x)$.  We call $q$ a quadratic form on $M$.
Let the Clifford algebra $\mathrm{Cl}(M,q)$ be defined in the usual way (e.g., as the quotient algebra $T(M)/I_q$, where $I_q$ is the ideal generated by elements of the form $x\otimes x-q(x)$, etc.)
Question 1:  Is there any problem with this definition?  (I am aware that bimodules introduce potentially curious structure between left and right scalar multiplications, and thus this definition allows some new 'twisted' Clifford algebras even in the commutative case.  I do not see that this impedes the definition.)
Question 2:  Since the definition of a Clifford algebra appears to extend pretty naturally to a noncommutative base ring $R$ and bimodule $M$ with essentially no change from the commutative case, presumably this must be a fairly well-known in mathematics – unless I have overlooked some stumbling block.  Where might I find a reference at undergrad/postgrad level that explores this?  I have not found anything through online searches.
Question 3:  I am new to bimodules and I have not rigorously explored this, but it seems that a free bimodule (of finite rank) would have bases with respect to the left-module structure and bases with respect to the right-module structure, but a left-basis is not necessarily a right-basis.  In particular, I conjecture that such a bimodule can be decomposed into the direct sum of cyclic (rank-1) bimodules, which would allow a simpler analysis of bimodules.  Is this valid?
I can add some specific examples for my conjectures if this helps.
 A: A Clifford algebra over a noncommutative ring defined in this way is generally quite degenerate compared to what we would expect, and is consequently not likely to be of much interest.
By way of illustration, let a $\mathbf{H}$-$\mathbf{H}$-bimodule have both a left- and right-basis $\{e_0,e_1\}$ such that $r.e_0=e_0.r$ and $s.e_1=e_1.s$.  We can verify that this is consistent with the bimodule axioms.  Let $q$ be a quadratic form defined by $q(r.e_0+s.e_1)=r^2+s^2$.  Let $b$ be the bilinear form $b(r.e_0+s.e_1,t.e_0+u.e_1)=rt+su$.  We can verify that $b(r.e_0+s.e_1,r.e_0+s.e_1)=r^2+s^2=q(r.e_0+s.e_1)$.
We can verify that an algebra that satisfies the identity $(re_0+se_1)^2=q(r.e_0+s.e_1)$ has $e_0^2=e_1^2=1$, $e_0e_1+e_1e_0=0$, and that scalars commute with this basis.
Let $r$ and $s$ be two scalars such that $rs^2\ne s^2r$ and $sr^2\ne r^2s$ (such quaternions are easy to find), which implies $rs-sr\ne 0$.  It follows that $r^2+s^2=(re_0+se_1)^2=r^2e_0^2+rse_0e_1+sre_1e_0+s^2e_1^2=r^2+s^2+(rs-sr)e_0e_1$, and since every nonzero quaternion is a unit, it follows that $e_0e_1=0$.  We can further show that $(re_0+se_1)^2(re_0+se_1)=(r^3+s^2r)e_0+(r^2s+s^3)e_1$ and $(re_0+se_1)(re_0+se_1)^2=(r^3+rs^2)e_0+(sr^2+s^3)e_1$.  Since the coefficients of the basis are unequal, the algebra is either not even power-associative or the bimodule maps into the algebra via the zero homomorphism.
This shows that the construction as suggested in the question is not what one might have hoped; indeed, in this example (and I guess many others), it produces an uninteresting algebra.  As a quotient of the tensor algebra, which is inherently associative, this example will be isomorphic to $\mathbf{H}$.
This does not imply that Clifford algebras cannot be generalized usefully to noncommutative rings, only that the ideal for taking the algebra quotient must be defined differently.
