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.
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 $$e_1^2=5,\quad e_2^2=6,\quad e_1\cdot e_2=e_{12},\quad e_2\cdot e_1=2-e_{12}.$$