# Clifford algebra of a non-diagonal quadratic form over rings

I know how to construct explicitly the Clifford algebra of a quadratic form over fields, even in the case the diagonal quadratic form over rings. But how should I $$\textbf{construct explicitly}$$ the Clifford algebra of a non-diagonal quadratic form over rings, for example, what is the associated Clifford algebra to the $$5x^2+2xy+6y^2$$?

I mean the $$\textbf{explicit construction}$$.

Thank you.

## 1 Answer

I'm not sure what you mean by "explicit", since the definition is already pretty explicit: if $$(M,q)$$ is a quadratic module over a commutative ring $$R$$, then $$C(M,q)$$ is the quotient of the tensor algebra $$T(M)=\bigoplus_{n\in \mathbb{N}} M^{\otimes n}$$ by the ideal generated by all $$x\otimes x - q(x)$$ with $$x\in M$$. In particular it is generated as a $$R$$-algebra by $$R=M^{\otimes 0}$$ and $$M=M^{\otimes 1}$$. By definition, for any $$x,y\in M$$, we have $$x^2=q(x),$$ and since $$(x+y)^2=q(x+y) = q(x) + b_q(x,y) + q(y)$$ where $$b_q$$ is the polar form of $$q$$, we find $$xy + yx = b_q(x,y)$$ which tells us how elements of $$M$$ commute in $$C(M,q)$$.

In your first example, $$M$$ is a free module of rank 2, say $$M=Re_1\oplus Re_2$$, and the quadratic form $$q$$ satisfies $$q(e_1) = 5,\quad q(e_2)=6,\quad b_q(e_1,e_2)=2.$$ So $$C(M,q)$$ will be a free $$R$$-module, with basis $$(1,e_1,e_2,e_{12})$$, and the product is determined by $$e_1^2=5,\quad e_2^2=6,\quad e_1\cdot e_2=e_{12},\quad e_2\cdot e_1=2-e_{12}.$$