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Let $A$ be a commutative euclidean ring, (probably) with 2 a unit in $A$. I'm trying to compute Witt and Grothendieck-Witt rings and since $A$ is a PID any f.g.p. module over it is free so I only need think about forms on $A^n$.

Question: Is every (non-degenerate) quadratic form over $A$ diagonalizable?

A form $q$ is diagonalizable we can perform a base change on $A^n$ such that the matrix for $q$ becomes diagonal. Edit: Non-degenerate here means that any matrix associated to the form is invertible.

From Milnor-Husemoller's book I know this is true if $A$ is local. If I can show that any non-degenerate quadratic form on $A^n$ represents some unit ($q(x) = u$ a unit for some $x \in A^n$) then the statement holds by cor. I.3.3 in Mil-Hus.

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What do you mean by "non-degenerate"? The condition that "any non-degenerate quadratic form on $A^n$ represents some unit" is a bit hard to satisfy if forms like $p\sum_i x_i^2$ are allowed... – darij grinberg Aug 20 '10 at 18:29
If you are looking for the largest class of rings where things that hold for quadratic forms over fields still mostly work, maybe the best you can do is semilocal rings like in Ricardo Baeza's book. Characteristic 2 plays the usual havoc. – Skip Aug 20 '10 at 18:39
@darij: I edited, so now I'm not allowing those kinds of forms. @Skip: Thanks I'll check that out. – K.J. Moi Aug 20 '10 at 18:44
I did not initially realize this question was in my area. It appears you need to consider easier things before returning to this. I always recommend $$ $$ The Arithmetic Theory of Quadratic Forms, by Burton Wadsworth Jones $$ $$ Integral Quadratic Forms, by George Leo Watson $$ $$ Rational Quadratic Forms, by John William Scott Cassels $$ $$ and then back to your area, the recent $$ $$ Basic Quadratic Forms, by Larry J. Gerstein, where Gerstein, as well as T. Y. Lam, can say ``Pfister'' without snickering. – Will Jagy Aug 20 '10 at 21:13
up vote 2 down vote accepted

Quadratic forms over $\mathbb{Z}$ don't diagonalize in general. Even positive definite rank two forms like $3x^2+2xy+5y^2$ can't be diagonalized. Inverting $2$ won't help things.

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Will demanding that the determinant of the corresponding matrix (when 2 is invertible) help? – K.J. Moi Aug 20 '10 at 18:45
Not much. Look at "unimodular lattices". Admittedly the earlier nontrivial examples can be diagonalized when you inverrt $2$ but eventually that trick won't work any more. – Robin Chapman Aug 20 '10 at 19:02

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