# Clifford Algebra and quadratic forms

I was wondering if someone can help me with the following. The question is,

How can I show that $C_0(q) = R[t]/(t^2 - bt + ac)$, given that $q(x, y) = ax^2 + bxy + cy^2$ is a quadratic form in two variables over the Ring $R$ with $a, b, c \in R$?

I am trying to study the general technique in doing problems like this. Thanks!

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Let us first describe the ring $\mathbb R[t] / p$ where $p = t^2 - bt + ac$ is any quadratic polynomial over $\mathbb R$. If its discriminant $D = b^2 - 4ac$ is negative then $p$ is irreducible and $R[t] / p \cong \mathbb C$. If the discriminant is positive then $p$ factors over $\mathbb R$ as $p = (t-a)(t-b)$ and we will have $\mathbb R[t] / p \cong \mathbb R \oplus \mathbb R$ with zero divisors but nontrivial Clifford product. Finally, for $D = 0$ we have a repeated root and so the quotient will also contain nilpotent elements. This is simply a Grassman algebra with $e \wedge e = 0$ where I've written $e$ for the equivalence class of $(t-c)$ in the quotient, $c$ being a double-root of $p$.
Now, to the quadratic form $q$ we have associated a (possibly degenerate) inner product $2\langle u,v\rangle = q(u+v) - q(u) - q(v)$, $u, v \in \mathbb R^2$, which can be represented in the basis $\{x, y\}$ by a matrix $$A = \pmatrix{a & b/2 \cr b/2 & c}.$$ Now, observe that $\det A = ac - b^2/4$ and so the properties of the polynomial $p$ are tight together with the signature of the quadratic form $q$.
Thank you for taking time looking at this. However, is there a way to approach this by considering first $q : M \rightarrow N$, where $M, N$ are $R$-modules, and then specifically treating $M = eR \oplus fR$? This would make $1, e, f, ef$ a basis for $C(q)$. Thanks again... – Eric Jan 15 '13 at 12:42
Sir @Marek, isn't that alright, since we are only looking for $C_0(q)$, which is two-dimensional? Please forgive my ignorance over this matter... – Eric Jan 15 '13 at 12:59
I should add that along with $M = eR \oplus fR$, we take $q(xe + yf) = ax^2 + bxy + cy^2$. Thanks! – Eric Jan 15 '13 at 13:01