Where does the theory of quadratic forms fail in characteristic 2?

Question

Let $V$ be a finite-dimensional vector space over a field $k$, and $Q$ a nondegenerate quadratic form on $V$. If the characteristic of $k$ is not 2, then we can change coördinates on $V$ so that $Q(\vec{x})=\sum_ia_ix_i^2$ for some $a_i\in k.$ Thus we can decompose $V$ as $(V,Q)=\oplus(V_i,Q_i),$ where the $V_i$ are $1$-dimensional and $Q_i(x)=a_ix^2.$

My question is: can we do something similar in the case where $char(k)=2$? If not, why not, and what can we do instead?

Answer 1

As to what can be done instead, I quite like this book, GROVE. The book you will see most on fields rather than rings is LAM. Also quite modern, but probably a bit easier to digest, is GERSTEIN.

Grove has the most detail, three chapters entirely on characteristic 2, one of those on Clifford Algebras but the other two on Orthogonal Groups/Geometry, meaning quadratic forms.

Answer 2

Consider the quadratic form $q(x, y) = xy$. Over a field of characteristic not equal to $2$, we can use the change of variables $x = u + v, y = u - v$ and so write this quadratic form in the variables $u, v$ as $q(u, v) = u^2 - v^2$. Then $u = \frac{x + y}{2}, v = \frac{x - y}{2}$. Over a field of characteristic $2$, we can no longer divide by $2$, so this change of variables is no longer invertible. In fact, it's easy to see that starting from any diagonal form $\sum a_i x_i^2$ and changing coordinates, all of the off-diagonal terms $x_i x_j, i \neq j$ vanish (because replacing $x_i = \sum b_j y_j$ gives $x_i^2 = (\sum b_j y_j)^2 = \sum b_j^2 y_j^2$); in other words, the diagonalizable forms are precisely the diagonal forms.

You can find some additional references at this MO question.