Meaning of pseudo-discriminant of a quadratic form

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.

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.

• Thanks a lot.!! – justanothermathstudent Sep 11 at 12:00
• I've added some comments about this notion in my answer. Hope this helps. – GreginGre Sep 11 at 12:02
• Thanks again, sorry for late reply. Doesn't the existence of symplectic basis require the characteristic to be not equal to 2? – justanothermathstudent Sep 12 at 14:08
• No I was mistaken sorry. – justanothermathstudent Sep 12 at 14:10