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Let $E$ be an elliptic curve over field $\mathbb{Z}/p\mathbb{Z}$.

The curve group $E(\mathbb{Z}/p\mathbb{Z})$ is always a) cyclic or b) direct product of two cyclic groups.

First question: How do I tell for a given field and curve if it's the case a) or b) ?

Another question follows from an example: The curve $E: y^2 = x^3 - x$ over $\mathbb{Z}/71\mathbb{Z}$ has 71 affine points and a point at infinity so it's group is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/36\mathbb{Z}$ (that's the case b)).

Second question is: Why it's not isomorphic to $\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/24\mathbb{Z}$ or $\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/18\mathbb{Z}$ ?

In other words, for b) case curves, how do I tell which are those two in the product?

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  • $\begingroup$ Dear Number Four, two questions: (1) What is your definition of an elliptic curve over $\mathbf Z/n\mathbf Z$? (2) What is your definition of the group of its points with coefficients in $\mathbf Z/n\mathbf Z$? $\endgroup$ Nov 27, 2013 at 23:12
  • $\begingroup$ I kind of updated the question - the definition of a curve over a ring would possibly make sense only for a commutative ring. Anyway I work with Weierstrass nonsingular curves over a finite field. $\endgroup$
    – NumberFour
    Nov 28, 2013 at 0:00
  • $\begingroup$ $\mathbf Z/n\mathbf Z$ is commutative, so that wasn't really a problem. But I don't understand how your example relates to the question, because all three groups that you have written down are products of two cyclic groups. Also, a group can be both cyclic and the product of two nontrivial cyclic groups (Chinese Remainder Theorem). $\endgroup$ Nov 28, 2013 at 0:08
  • $\begingroup$ Yeah, but I meant the general definition would need to have word 'commutative'. My point in the example is why isomorphic to $\mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/36\mathbf{Z}$ and not to $\mathbf{Z}/3\mathbf{Z} \times \mathbf{Z}/24\mathbf{Z}$ or the others? $\endgroup$
    – NumberFour
    Nov 28, 2013 at 0:13
  • $\begingroup$ That is also a legitimate question, but what I am saying is that I don't understand how it relates to the question just above it, of which it is supposed to be an example. $\endgroup$ Nov 28, 2013 at 0:14

1 Answer 1

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For simplicity, I’ll write $C_n$ for $\mathbb Z/n\mathbb Z$.

First, I think you’ll find that $C_2\times C_{36}$ and $C_4\times C_{18}$ are isomorphic. The best way to write something like this is to make sure the indices divide: $2|36$, but $4$ does not divide $18$.

Your question about why $C_3\times C_{24}$ does not occur is much more interesting, and I’m almost out in the water over my head in attempting an answer. There’s a fancy gadget called the $e_n$-pairing on the points of order $n$ of an elliptic curve, and it’s defined over the ground field, with values in the multiplicative group scheme $\mathbf G_{\mathrm m}$. The upshot is that if all $n^2$ points of the elliptic curve of period $n$ are $k$-rational ($k$ being the ground field), then the $n$-th roots of unity have to be in $k$ as well. But of course $\mathbb F_{71}$ doesn’t have cube roots of unity. (I know that I’ve left out far too many details here, but the topic is fairly advanced, and I don’t have the skills.)

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  • $\begingroup$ Thanks for your reply. Firstly, why the curve group from the example is not isomorphic to $\mathbf{Z}/72\mathbf{Z}$? Then it would be(through Chinese remainder theorem) isomorphic to both $C_2 \times C_{36}$ and $C_3 \times C_{24}$, wouldn't it? $\endgroup$
    – NumberFour
    Nov 28, 2013 at 14:56
  • $\begingroup$ No, neither of those, since neither of those has an element of period $72$. Yet $C_{72}$ is isomorphic to $C_8\times C_9$. $\endgroup$
    – Lubin
    Nov 28, 2013 at 15:15
  • $\begingroup$ @NumberFour: Check the hypotheses of the Chinese Remainder Theorem. $\endgroup$
    – Alex B.
    Nov 30, 2013 at 14:09
  • $\begingroup$ Yeah, I completely forgot - $C_2 \times C_{36}$ neither is $C_3 \times C_{24}$ are isomorphic to $C_{72}$, because neither 2 and 36 nor 3 and 24 are coprime. Anyway, I think I found the reason why the elliptic group from the example isn't isomorphic to $C_3 \times C_{24}$: The elliptic group doesn't have an element of order 24, only elements of orders 1,2,3,4,6,9,12,18,36, thus $C_2 \times C_{36}$ is the only option. $\endgroup$
    – NumberFour
    Nov 30, 2013 at 15:03
  • $\begingroup$ An answer that is not an answer...? $\endgroup$ Sep 1, 2020 at 7:18

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