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(1) I am currently trying to find out how to classify all elliptic curves over an arbitrary field. Several questions have accumulated while I was trying to answer this question. If you don't want to read the whole question, only read part (3) or (4). A satisfying answer to one of these would solve my problem.

(2) Can elliptic curve over imperfect field be reduced to Weierstrass form? This old question has never been answered, so I'm including it here. The easy definition of the j-invariant always uses the weierstrass-form. If the curve cannot be converted into weierstrass-form, how is the j-invariant defined in the first place? Is there a general way to do it that's independent of field and form?

(3) It seems like twists are the way to classify curves that happen to be isomorphic over the closure (or the seperable closure?) of a given (arbitrary) field. In this thread Mr. Silverman does a great way to explain how this works and even how to define an invariant from that method: J-invariant and isomorphism of elliptic curves over $\mathbb{Q}$ Is it possible to extend/ modify Mr. Silvermans answer, so that the classification works in arbitrary fields?

(4) I have tried to find an answer in the literature. The classification from Mr. Silvermans answer above is done in both his and in Mr. Husemöllers (Chapter 7) books, but, as far as I understand it, both only classify curves for perfect fields. It seems like something similar is possible for arbitrary fields, though. In the book "Rational Points on Varieties" by Poonen, Chapter 4, a theorem is included for twists of quasi-projective k-varieties. This classifies curves that become isomorphic over an arbitrary Galois extension and thus also over the separable closure. This is only useful, though, if we can classify the curves over the separable closure of a given field. As far as I understand it, to classify all the different curves that are isomorphic over the separable closure of a field would require some different method because the j-invariant will only work if the separable closure is equal to the algebraic closure (case of an perfect field). How do you classify all curves over the separable closure of a non perfect field? Secondly, this book is a little bit hard. To be honest, at the moment, I don't even understand enough scheme theory do verify with certainty that elliptic curves are quasi-projective varieties in the scheme-theoretic sense. Are they? Furthermore, the book is pretty technical. If this is the way to do the classification in the case of (truly) arbitrary fields, then my last question is, wether there is a less technical and maybe less general (as I only want to do it for elliptic curves) way to do it? Do you know of any good reference for this?

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    $\begingroup$ Let me know if the fact below answers your question. I suspect the method doesn't make you happy, since you say you don't know scheme theory, but I'm not sure it's possible (or at least short) to put it in non-scheme theory terms. $\endgroup$ Jun 6, 2021 at 18:15
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    $\begingroup$ Hello: I can definitely see how this was closed as "needs focus" in the past. When authoring posts, please attempt to isolate one, or possibly two closely related questions. When you have three or four, as you seem to have, (It's hard to even count for sure.) it makes it hard to read and hard to give a complete solution to. It would be totally fine (advisable even) to do them in separate posts. If it is the case that one of the items remains unsolved, please edit to move it to a post of its own and remove it from this one. $\endgroup$
    – rschwieb
    Jun 28, 2021 at 20:44
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    $\begingroup$ @rschwieb thank you for your feedback. I already got two answers below that helped to point me in the right direction. I realised that the question is not focused enough when it was shortly closed. I didn't change it for now because I already accepted one of the answers. If I need to ask another question, though, I will make sure to keep it shorter and more focused. $\endgroup$
    – Guenterino
    Jun 29, 2021 at 8:25

3 Answers 3

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This is too long for a comment. If it doesn't answer your question, please let me know. I also wrote this in a rush, so please double check my claims.

I don't know if you are focused on characteristic 2 or 3, but I think if $k$ is of characteristic $p>3$ then for elliptic curves $E_i$ for $i=1,2$ over $k$ one has that $(E_1)_{\overline{k}}\cong (E_2)_{\overline{k}}$ iff $(E_1)_{k^\mathrm{sep}}\cong (E_2)_{k^\mathrm{sep}}$. The reason is 'simple' (but uses heavy machinery). Consider the functor

$$\mathrm{Isom}(E_1,E_2):\mathbf{Sch}_{/k}\to\mathbf{Set},\qquad T\mapsto \mathrm{Isom}_{\mathbf{Sch}_{/T}}((E_1)_T,(E_2)_T)$$

If $E_1$ and $E_2$ are isomorphic over $\overline{k}$ then, in particular, it's not hard to see that $\mathrm{Isom}(E_1,E_2)$ is a fpqc torsor for $\mathrm{Aut}(E_1)$. This is a finite group scheme over $\mathrm{Spec}(k)$, and since affine morphisms satisfy fpqc descent (see Tag 0245) we know that the $\mathrm{Aut}(E_1)$-torsor $\mathrm{Isom}(E_1,E_2)$ is representable by some $k$-scheme $I$. But, since $\mathrm{char}(k)>3$ we know that $|\mathrm{Aut}(E_1)|$ is invertible in $k$ (cf. [Silv, Theorem 10.1]) and so we know that $\mathrm{Aut}(E_1)$ is smooth over $\mathrm{Spec}(k)$ (e.g. see [Mil, Corollary 11.31]), and so $I$ must also be smooth over $\mathrm{Spec}(k)$ (e.g. see Tag 02VL). But, then $I(k^\mathrm{sep})\ne\varnothing$ (see [Poon, Proposition 3.5.70]). Thus, $(E_1)_{k^\mathrm{sep}}\cong (E_2)_{k^\mathrm{sep}}$.

So, asuming $\mathrm{char}(k)>3$ one can use usual Galois descent arguments to classify elliptic curves over $k$ (e.g. see Theorem 27 of my blog post here).

For characteristic $p=2,3$, the same thing works for classifying forms of $E$ assuming that $p\nmid |\mathrm{Aut}(E)|$. If $p\mid |\mathrm{Aut}(E)|$ I'm not sure what happens.

References

[Mil] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

[Poon] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..

[Silv] Silverman, J.H., 2009. The arithmetic of elliptic curves (Vol. 106). Springer Science & Business Media.

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    $\begingroup$ @Guenterino I haven't checked anon's answer below, but it seems plausible, so heavy machinery may be avoidable (even if the more elementary approach is minorly ad hoc). It is true that [Poon, Theorem 4.5.2] covers elliptic curves, but there it is talking only about twists as varieties--not counting the group structure. I am secretly hiding here that one can do this also for the group structure (not hard with some machinery--let me know if you want me to explain), and so I am using the automorphisms as a GROUP VARIETY opposed to just an abstract variety (which is what Poonen is doing). $\endgroup$ Jun 6, 2021 at 23:33
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    $\begingroup$ So yes, my automorphism group/Isom varieties take 0 to 0. The issue with characteristic 2 and 3 can be seen in different ways. In anon's answer it's about not being able to put things in standard Weierstrass form. In my answer it's about the fact that my $\mathrm{Aut}(E)$ is not 'smooth'. Like what is being used here, in short, are the following two facts. a) the 'moduli space of isomorphisms between two elliptic curves' is representable and automatically smooth in characteristic larger than 3, b) that smooth varieties which are non-empty (i.e. have non-empty set of $\overline{k}$ points) $\endgroup$ Jun 6, 2021 at 23:36
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    $\begingroup$ automatically have a non-empty set of $k^\mathrm{sep}$-points. So, the issue for characteristic 2 and 3 is that a) can be false. For instance, $\mathrm{Isom}(E_1,E_2)$ is smooth if and only if $\mathrm{Aut}(E_1)$ is smooth. But, $\mathrm{Aut}(E_1)$ can be $\mu_2$ which is not smoot in characteristic $2$. In general, if $p$ (the characteristic of $k$) divides $|\mathrm{Aut}(E)|$ then $\mathrm{Aut}(E)$ isn't smooth. By the classification of what group (schemes) $\mathrm{Aut}(E)$ can be, it can only be divisible by the primes $2$ and $3$. So in characteristic greater than $3$ there's no problem. $\endgroup$ Jun 6, 2021 at 23:39
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    $\begingroup$ @Guenterino The high level explanation is given in Lemma 24 of my blog post. Given what I said above, if characteristic of $k$ is greater than $3$, one can replace $\overline{k}$ with $k^\mathrm{sep}$. $\endgroup$ Jun 7, 2021 at 13:01
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    $\begingroup$ But I wonder if it is necessary to use that heavy machinery. In Qing Liu's book, Riemann--Roch theorem is stated over an arbitrary field, and exercise 4.1 in chap. 7 asks to prove that if X is a smooth, geometrically connected, projective curve of genus 1 over a field k, with a k-rational point, then it defines a Weierstrass equation, just using Riemann--Roch. A similar treatment can be found as (1.8) Example in Moonen--Geer--Edixhoven notes on abelian varieties. $\endgroup$
    – Watson
    Jun 8, 2021 at 6:51
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It depends what you mean by Weierstrass form. Every elliptic curve over a field $k$ is described by an equation $Y^{2}Z+a_{1}XYZ+a_{3}YZ^{2}=X^{3}+a_{2}X^{2}Z+a_{4}XZ^{2}+a_{6}Z^{3}$. See M p47.

If the characteristic is not 2 or 3, then an elementary argument (M p50) shows that you can put this in the form $E(a,b):Y^{2}Z=X^{3}+aXZ^{2}+bZ^{3}$. If the characteristic is 2 or 3, you may not be able to put it in this form.

If the characteristic is not 2 and 3, it is easy to see that two curves E(a,b) and E(a',b') become isomorphic over a separable extension of $k$ if they have the same $j$-invariant (Proof of (c) of Theorem 2.1, M p51). Indeed E(0,b) and E(0,b') become isomorphic over the field obtained by adjoining a sixth root of b/b' and E(a,b) and E(a',b') become isomorphic over the field obtained by adjoining a 4th root of a/a' and a square root of -1. If you avoid 2 and 3, everything works much as in characteristic zero, except you replace the algebraic closure of $k$ with the separable closure.

In characteristic 2 and 3, you have to work with more complicated equations and I don't know whether two curves with the same $j$ invariant become isomorphic over a separable extension of the base field.

You can find all of this in Milne's or Silverman's books on Elliptic Curves. For example, the definition of the j-invariant of an elliptic curve in general Weierstrass form can be found on M p53.

M= first edition of Milne's book, available on his webpage (under Books).

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  • $\begingroup$ Thank you very much for your answer. $\endgroup$
    – Guenterino
    Jun 7, 2021 at 8:02
  • $\begingroup$ If I understand it correctly, though, Milne introduces the j-invariant by using the normal form E(a,b). If this form does not always exist for characteristic 2 and 3, do you know how the j-invariant is defined in these cases? $\endgroup$
    – Guenterino
    Jun 7, 2021 at 8:07
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    $\begingroup$ Many many years ago, Tate gave formulas for using the general (5-parameter) W-form to get the $j$-invariant. These must be available somewhere; look in Silverman. If you can’t find it at all, I’ll send you a copy. $\endgroup$
    – Lubin
    Jun 7, 2021 at 20:12
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The paper

Michael Szydlo, Elliptic fibers over non-perfect residue fields, J. Number Theory 104 (2004), no. 1, 75-99 (MR2021627)

is a detailed study of the possible reduction types of elliptic curves over non-perfect fields.

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