Timeline for Semistability of Frey curves: why no additive reduction?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 5 at 7:55 | comment | added | Chris Wuthrich | The answer to your "Is it true?" question is "No". Take any Weierstrass equation $y^2=x^3+Ax+B$ with integers $A$ and $B$. Now $Y^2 = X^3 +A\ell^4 x + B\ell^6$ is also a model whose reduction is clearly additive at $\ell$, yet the reduction type of the initial equation (which could be minimal at $\ell$) could be anything. | |
| Jun 5 at 7:52 | comment | added | Chris Wuthrich | If a Weierstrass equation with integer coefficients has multiplicative reduction at a prime $\ell$, then the denominator of the $j$-invariant is divisible by $\ell$. Therefore in any model with integer coefficients the reduction will be multiplicative or additive, and the additive case only arises when the model is not minimal at $\ell$. This explains Serre's remark. And your last question. | |
| Jun 5 at 7:05 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| S Jun 5 at 5:34 | review | First questions | |||
| Jun 5 at 5:44 | |||||
| S Jun 5 at 5:34 | history | asked | Johnny Apple | CC BY-SA 4.0 |