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
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| May 31, 2017 at 3:57 | history | edited | Stanley Yao Xiao | CC BY-SA 3.0 |
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| Aug 11, 2011 at 16:06 | comment | added | Daniel Erman | @Joe :) I agree it's useful to leave this answer. | |
| Aug 11, 2011 at 15:02 | history | edited | JSE | CC BY-SA 3.0 |
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| Aug 11, 2011 at 14:44 | comment | added | Joe Silverman | @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). | |
| Aug 11, 2011 at 14:25 | comment | added | Daniel Erman | 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 $s^2+ut^2$. | |
| Aug 11, 2011 at 14:18 | history | answered | Joe Silverman | CC BY-SA 3.0 |