Meaning of pseudo-discriminant of a quadratic form

Question

In Bourbaki Lie Groups and Lie algebras chapter 6 section 4 excercise 1(c), they have used the word pseudo-discriminant. The reference is given to be Algebra chapter IX which I can't find a English translation of. Here the following definition is given. Since I don't know French I can't understand the definition. Any help will be appreciated. Sorry if this kind of question is not for this site.

Answer 1

Here is a translation in English. I added few things into parentheses.

Assume that $A$ is a field of characteristic $2$, $E$ is a vector space on $A$ of even dimension $2r$, and $Q$ is a non-degenerate quadratic form on $E$.

a) Let $(e_i)$ a symplectic basis of $E$ for the alternating bilinear fomr $\Phi$ associated to $Q$ . Show that the element $z=e_1e_1+e_3e_4+\cdots+e_{2r-1}e_{2r}\in C^+(Q)$ (the even Clifford algebra) forms, together with the unit element, a basis of the center $Z$ of $C(Q) (the Clifford algbra).

Moreover, $Z$ is the direct sum of two fields of and only if th element $\Delta(Q)=Q(e_1)Q(e_2)+Q(e_3)Q(e_4)+\cdots+ Q(e_{2r-1})Q(e_{2r})$, called the pseudo-discriminant of $Q$ with respect to the symplectic basis $(e_i)$, has the form $\lambda^2+\lambda$ for some $\lambda\in A$.

Some comments on this notion. Even if this is not explicitely asked by the OP, I think it would be nice to comment on this notion.

Recall that the determinant $\det(Q)$ of a non degenerate quadratic form $Q$ over a field $A$ (to stick with the previous notation) is the square class of the determinant of any representative matrix of the associated bilinear form. It is an element $A^\times/A^{\times 2}$, which is an invariant of the isometry class of $Q$ (two isometric quadratic forms have equal determinants). If $A$ has characteristic different from two, this invariant is quite useful, and may be use to obtain classification results.

If $A$ has characteristic two, then a representative matrix of the associated bilinear form is alternating, so its determinant is always a square. Hence, in this case , $\det(Q)$ is always the trivial square class.

Set $\wp(A)=\{ \lambda^2+\lambda\mid \lambda\in A\}$. This is a subgroup of $A$. One may show that the class of $\Delta(Q)$ in $A/\wp(A)$ does not depend on the choice of the symplectic basis, so we can set ${\rm Arf}(Q)$ to be the class of $\Delta(Q)$ in $A/\wp(A)$. The class ${\rm Arf}(Q)$ is called the Arf invariant of $Q$. One may show this is an anvariant of the isometry class of $Q$. It plays the same role as the determinant.