Timeline for If the discriminant of a binary quadratic form has high valuation, is the form "almost a square".
Current License: CC BY-SA 3.0
21 events
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| Aug 24, 2011 at 16:16 | comment | added | Melanie Matchett Wood | Example: $a=t$ and $b=2us+2t$ and $c=2us+t$. The discriminant is $4u^2s^2$. This form is not congruent to a square modulo $u^2$ | |
| Aug 18, 2011 at 1:59 | comment | added | Daniel Erman | @JSE's last comment: I guess this is correct! At the moment, I don't see how to construct another example, so I'll have to think more about your second comment in this thread. :) | |
| Aug 14, 2011 at 2:21 | comment | added | JSE | Yeah, sorry. Dan's form is congruent mod u^2 to the square of (s+ut)x + ty right? | |
| Aug 14, 2011 at 2:18 | comment | added | JSE | Um, no, wait a minute. | |
| Aug 14, 2011 at 2:06 | comment | added | JSE | Of course, in the real case, you can "drop a perpendicular" from f to the disc-0 surface, and get an L^2 within disc(f) of f. That works fine for dvrs and must be the same as completing the square as Joe does. | |
| Aug 14, 2011 at 2:04 | comment | added | JSE | Comment: suppose I take a quadratic form f which I know is within epsilon of a perfect square L^2. (You can think of eps as 1/p here.) How small is disc(f)? Well, we can think of f as what you get when you start from L^2 and move an infinitesimal distance eps in some direction. If that direction is not tangent to the disc-0 surface, disc(f) is on order eps. If the direction IS tangent to the disc-0 surface, disc(f) is on order eps^2. But it cannot be any more because the disc-0 surface has no lines of contact order 3. (I am ignoring the point L=0 which is different somehow.) | |
| Aug 14, 2011 at 1:59 | comment | added | JSE | Got it, and verified that Dan's example is not congruent mod u^2 to the square of any linear form. | |
| Aug 11, 2011 at 19:02 | comment | added | Melanie Matchett Wood | If we take Jordan's example $a = s^2 + pust$, $b = 2st + put^2$, $c = t^2$, then the discriminant is $p^2u^2t^4$, which has $u$-adic valuation $2$ (which is the relevant one---the $2$ or $p$ is a red herring). | |
| Aug 11, 2011 at 16:26 | comment | added | JSE | But is this just a 2-ological phenomenon? If a = s^2 + pust, b = 2st + put^2, c = t^2 I think the discriminant has p-adic valuation 1, not 2. | |
| Aug 11, 2011 at 15:24 | answer | added | JSE | timeline score: 2 | |
| Aug 11, 2011 at 15:22 | comment | added | Daniel Erman | @JSE I'm not sure exactly what example you want (in the 2nd comment), but here's an attempt. Set $R=\mathbb Q[[u]]$, and let $a=s^2+2ust$, $b=2st+2ut^2$ and $c=t^2$. Then, if I've done this correctly, $b^2-4ac=4u^2t^4$. Is this what you were looking for? | |
| Aug 11, 2011 at 15:15 | comment | added | JSE | indeedieweedie. | |
| Aug 11, 2011 at 15:02 | comment | added | Emerton | Dear Jordan, Do you rather mean the condition that one of $a, b, c$ be a unit? Regards, Matt | |
| Aug 11, 2011 at 14:45 | comment | added | JSE | (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.) | |
| Aug 11, 2011 at 14:41 | comment | added | JSE | 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. | |
| Aug 11, 2011 at 14:38 | history | edited | Daniel Erman | CC BY-SA 3.0 |
Added clarification and an example.
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| Aug 11, 2011 at 14:18 | answer | added | Joe Silverman | timeline score: 2 | |
| Aug 11, 2011 at 14:15 | comment | added | Daniel Erman | @Joe Silverman. Thanks! I just removed the mistaken 2's. | |
| Aug 11, 2011 at 14:14 | history | edited | Daniel Erman | CC BY-SA 3.0 |
Typos: unnecessary 2's.
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| Aug 11, 2011 at 14:00 | comment | added | Joe Silverman | 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 $2$. | |
| Aug 11, 2011 at 13:37 | history | asked | Daniel Erman | CC BY-SA 3.0 |