2 clarifying the meaning of "non-degenerate"

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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# Diagonalization of quadratic forms over euclidean rings

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.

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.