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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.

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    $\begingroup$ In your claim, you either want the discriminant to be $b^2-ac$ or you want your quadratic form to be $ax^2+bxy+cy^2$. Probably you didn't mean to put the $2$ coefficient on the $b$, since your earlier $f$ doesn't have the $2$. $\endgroup$
    – Joe Silverman
    Commented Aug 11, 2011 at 14:00
  • $\begingroup$ @Joe Silverman. Thanks! I just removed the mistaken 2's. $\endgroup$
    – Daniel Erman
    Commented Aug 11, 2011 at 14:15
  • $\begingroup$ It might help to give an example where the discriminant has valuation nu and the largest valuation you can give a',b',c' is nu/2. $\endgroup$
    – JSE
    Commented Aug 11, 2011 at 14:41
  • $\begingroup$ (ok, now you've done so.) Do you have an example where the form is not within u^{nu/2} of 0? The question is in some sense about the differential of b^2-4ac near its vanishing locus, and this is different near the singular point (0,0,0) than it is anywhere else. (Note that the condition of not being close to the zero form is, in the dvr case, exactly Joe's condition that one of a,b,c be a non-unit.) $\endgroup$
    – JSE
    Commented Aug 11, 2011 at 14:45
  • $\begingroup$ Dear Jordan, Do you rather mean the condition that one of $a, b, c$ be a unit? Regards, Matt $\endgroup$
    – Emerton
    Commented Aug 11, 2011 at 15:02

2 Answers 2

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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.

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  • $\begingroup$ Nice counterexample to the rough version. $\endgroup$
    – Daniel Erman
    Commented Aug 11, 2011 at 16:11
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Factoring out whichever of $a,b,c$ has the smallest valuation, you can assume (I think) that at least one of $a,b,c$ is a unit. If $a$ is a unit, then $$ ax^2+bxy+cy^2 = a\left(x+\frac{b}{2a}y\right)^2 - \left(\frac{b^2-4ac}{4a}\right)y^2. $$ Similarly if $c$ is a unit. If $a$ and $c$ are non-units and $b$ is a unit, then the "$a$" coefficient of $f(x+y,y)$ is a unit. This seems to give your claim with the stronger result that $a'$, $b'$ and $c'$ all have valuation at least $\nu$. (Maybe I'm missing some subtlety here about general valuation rings?)

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  • $\begingroup$ I think this works when $Q=R$. But $Q$ is a polynomial ring over a valuation ring and is not itself a valuation ring (in particular it is not local). So if $a$ has valuation 0, then this does not imply that a is a unit. For instance, if $R=\mathbb Q[[u]]$, then a could be something like $s^2+ut^2$. $\endgroup$
    – Daniel Erman
    Commented Aug 11, 2011 at 14:25
  • $\begingroup$ @Daniel: Sorry. I knew I must be missing something (you wouldn't have asked such an easy question!). But maybe it's useful to leave my "answer" to indicate why the result is trivial for a valuation ring (as opposed to a ring that happens to have a valuation). $\endgroup$
    – Joe Silverman
    Commented Aug 11, 2011 at 14:44
  • $\begingroup$ @Joe :) I agree it's useful to leave this answer. $\endgroup$
    – Daniel Erman
    Commented Aug 11, 2011 at 16:06

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