Timeline for What is the minimal model of $E:y^2=x^3-x-n$?
Current License: CC BY-SA 4.0
9 events
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| Feb 7, 2021 at 0:01 | comment | added | Chris Wuthrich | My answer for $p>3$ and user171793's answer for $2$ and $3$ show that the equation is minimal. It is a consequence of Tate's algorithm that if $v(\Delta)<12$ then the equation is minimal at $v$; that is all that is needed from the algorithm. You may still want to learn about it - as explained in Silverman's second book. | |
| Feb 6, 2021 at 23:22 | comment | added | Milo Moses | @ChrisWuthrich I don't quite understand your comment; Could you restate it as an answer? | |
| Feb 4, 2021 at 21:41 | history | edited | Milo Moses | CC BY-SA 4.0 |
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| Feb 4, 2021 at 16:02 | comment | added | Milo Moses | @ChrisWuthrich I removed the "obviously" portion because I realized that my "obvious" argument was flat out incorrect | |
| Feb 4, 2021 at 15:51 | comment | added | Milo Moses | @user171793 Thank you for your comment; could you write it up as an answer so I can upvote and accept it? | |
| Feb 4, 2021 at 15:50 | history | edited | Milo Moses | CC BY-SA 4.0 |
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| Feb 4, 2021 at 10:56 | comment | added | user171793 | It is minimal (indeed good) at 3, as the discriminant is $\Delta=-2^4(27n^2-4)$ so $v_3(\Delta)=0$. Similarly, $v_2(\Delta)\le 7$ implies minimality at 2. | |
| Feb 4, 2021 at 8:55 | comment | added | Chris Wuthrich | The sentence starting with "Obviously" is incorrect. If you want to change your equation with $p$, the term in front of $x$ would be divided by $p^4$. This means that your equation is minimal except maybe at $2$ and $3$. For those you have to look at Tate's algorithm. From experiments with small $n$, I guess that the answer will be that the equation is indeed globally minimal. | |
| Feb 4, 2021 at 5:32 | history | asked | Milo Moses | CC BY-SA 4.0 |