TITLE: If the discriminant of a binary quadratic form has high valuation, is the form "almost a square". 
QUESTION [6 upvotes]: For a binary quadratic form $ax^2+bxy+cy^2$ over a field (characteristic not 2), the discriminant $b^2-4ac$ is 0 if and only if that form is the square of a linear form.  I am curious about an analogue of this fact over a ring with a valuation.  
 Rough version of question.   Let
$$f=ax^2+bxy+cy^2$$
be a binary quadratic form over a ring $Q$ with a valuation.  If the valuation of the discriminant $b^2-4ac$ is large (so heuristically $b^2-4ac$ is ``almost 0''), does this imply that $f$ is "almost a square"?
 More precise version of question. Here's the precise case I am interested in.  Let $R$ be a valuation ring, where 2 is a unit.  Let $Q$ be the standard graded ring $Q=R[s,t]$ over $R$, i.e. $\text{deg}(s)=\text{deg}(t)=1$ and $Q_0=R$.  The ring $Q$ inherits a valuation from $R$.  Let $f$ as above, where $a$,$b$, and $c$ are all homogeneous elements of $Q$, and where $\text{deg}(a)+\text{deg}(c)=2\cdot \text{deg}(b).$
 Claim:   If $b^2-4ac$ has valuation at least $\nu$, then there exist $d,e,f$, and $a',b',c'$ in $Q$, such that
$$
ax^2+bxy+cy^2=d(ex+fy)^2 + (a'x^2+b'xy+c'y^2),
$$
where $a',b',$ and $c'$ all have valuation at least $\nu/2$.
 Update:   Let me add some clarification and an example.  First of all, note that $Q$ is not a valuation ring, since $Q=R[s,t]$.  So an element with valuation $0$ is not necessarily a unit.  For instance, $s^2$ is an element of $Q$ with valuation $0$ that is not a unit.  
Also, here's an example.  Let $R=\mathbb Q[[u]]$ and let $f=u^3x^2+u^3y^2$.  Then $b^2-4ac=-4u^6$, which has valuation $6$.  We can (trivially) write $f=(0)^2+f$ as the sum of a square and something with valuation $6/2=3$.  So I think that the $\nu/2$ bound in the claim is optimal.

REPLY [2 votes]: Speaking to the rough version, not the precise version, I think you will need some conditions on Q for the statement to be true.  For instance, suppose Q is $R[T,U,V]/(TU-V^2)$, and let f be the form
$T x^2 + 2 V xy + Uy^2$.
The discriminant of f is 0, but I don't think this can be expressed as a constant multiple of a square of a linear form, which gives a negative answer to the statement in the case where nu is infinite.