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Stream: theory: mathematics

Topic: elliptic curve examples

John Baez (Mar 07 2024 at 05:37):

One way to state the Modularity Theorem involves the L-function of an elliptic curve. But this L-function is usually defined in a rather mysterious way. I want to look at the definition in some very simple cases.

John Baez (Mar 07 2024 at 05:46):

Here's one version of the definition. Namely, for an elliptic curve $E$ defined over the rationals,

$L(E,s)=\prod_pL_p(E,s)^{-1}$

where $p$ ranges over primes, and for any prime $p$

$L_p(E,s)=\begin{array}{ll} (1-a_pp^{-s}+pp^{-2s}) & \text{if } p\nmid N \\ (1-a_pp^{-s}) & \text{if }p\mid N \text{ and } p^2 \nmid N \\ 1 & \text{if }p^2\mid N \end{array}$

John Baez (Mar 07 2024 at 05:51):

where $N$ is the conductor of $E$, which has its own unmotivated-looking definition.

John Baez (Mar 07 2024 at 05:51):

These definitions, copied from Wikipedia, are even worse than they need to be.

John Baez (Mar 07 2024 at 05:54):

I'll try to write down a more easily parsed definition.

John Baez (Mar 07 2024 at 06:02):

Here's one... it takes a lot of stuff hidden in the definition of "conductor" and makes it explicit, so while it looks worse than the above it's actually not.

John Baez (Mar 07 2024 at 06:02):

$L(E,s)=\prod_pL_p(E,s)^{-1}$

where $L_p(E,s)$ is defined in 4 cases.

John Baez (Mar 07 2024 at 06:08):

Let $a_p = p + 1 - N_p$ where $N_p$ is the number of points of $E$ defined over the field with $p$ elements. For this to make sense we need to have written our elliptic curve as a cubic equation with integer coefficients. $N_p$ is the number of solutions in the integers mod $p$, plus one for the point at infinity.

John Baez (Mar 07 2024 at 06:10):

Then:

1) $L_p(E,s) = 1 - a_p p^{-s} + p^{1-2s}$ if $E$ has good reduction at $p$, meaning that it's smooth as a curve defined over the field with $p$ elements.

John Baez (Mar 07 2024 at 06:26):

2) $L_p(E,s) = 1 - p^{-s}$ if $E$ has 'split multiplicative reduction' at $p$.

3) $L_p(E,s) = 1 + p^{-s}$ if $E$ has 'nonsplit multiplicative reduction' at $p$

4) $L_p(E,s) = 1$ if $E$ has 'additive reduction' at $p$.

John Baez (Mar 07 2024 at 06:30):

The last three are different ways of not having good reduction at $p$. I don't really understand them and this is why I'd like to see some very small examples, but here is what Anton Hilado has to say:

In the case that an elliptic curve has bad reduction at p, we say that it has additive reduction if there is only one tangent line at the singular point (we also say that the singular point is a cusp), for example in the case of the curve $y^{2}=x^{3}$, and we say that it has multiplicative reduction if there are two distinct tangent lines at the singular point (in this case we say that the singular point is a node), for example in the case of the curve $y^{2}=x^{3}-3x+2$. If the slope of these tangent lines are given by elements of the same field as the coefficients of the curve (in our case rational numbers), we say that it has split multiplicative reduction, otherwise, we say that it has nonsplit multiplicative reduction. We note that since we are working with finite fields, what we describe as “tangent lines” are objects that we must define “algebraically”, as we have done earlier when describing the notion of a curve being singular.

John Baez (Mar 07 2024 at 06:31):

I find some of this confusing (especially the part about slopes of curves being or not being in the same field as the coefficients of the curve), but this is exactly why I'd like to see examples.

John Baez (Mar 07 2024 at 06:34):

I hope there's a more high-powered abstract way to talk about these things that makes the $L$-function of a curve look like a bit less of a mess. I know such a way for primes of good reduction - it's in my paper Dirichlet species and the Hasse-Weil zeta function. But I want to figure out what's happening for primes of bad reduction.

John Baez (Mar 07 2024 at 07:00):

I'm hoping that the L-functions and modular forms database will help me with examples. They've got 3,824,372 elliptic curves over $\mathbb{Q}$, and they list the bad primes for each curve.

John Baez (Mar 07 2024 at 07:03):

Yes, it seems to help. For example the curve

$y^2 - y = x^3 - x^2$

has bad reduction only over $p = 11$, and there they say it has split multiplicative reduction.

John Baez (Mar 07 2024 at 07:04):

I picked that curve because it looks simple, but I'd rather find a simple one that has bad reduction over a smaller prime, because I don't want to think about $\sim$ 11 solutions.

Reid Barton (Mar 07 2024 at 18:59):

If we write $f(x,y) = y^2 - y - (x^3 - x^2)$ and solve the system

• $f(x,y)=0$,
• $0 = \partial f / \partial x = 2x - 3x^2$,
• $0 = \partial f / \partial y = 2y - 1$

over the field $\mathbb{F}_{11}$, we find a single nonsmooth point $(x = 8, y = 6)$. (There is a second point $(0, 6)$ where the partial derivatives are both zero, but it's not on the curve. I'm also ignoring the point at infinity which I think is always smooth for a curve given in Weierstrass form.)

We could replace $x-8$ by $x$ and $y-6$ by $y$ to move the non-smooth point to the origin, for convenience. By my calculation that produces a new formula $y^2 = x^3 + 3x^2$.

Now to find the tangent directions, we could try to solve this equation in power series $x$, $y \in k[[t]]$, where $k$ is some field extension of $\mathbb{F}_{11}$. Say

• $x = at + O(t^2)$,
• $y = bt + O(t^2)$.

If we plug these into $y^2 = x^3 + 3x^2$ we get

$$b^2 t^2 = 3 a^2 t^2 + O(t^3).$$

So we'll have a nontrivial solution $(a, b) \ne (0, 0)$ if and only if 3 is a square in $k$, say $3 = c^2$. Then there will be two solutions (up to rescaling) $(a, b) = (1, c)$ and $(a, b) = (1, -c)$, which are the two tangent directions.
In this case, $3 = 25 = 5^2$ is already a square in $\mathbb{F}_{11}$--so the curve has split multiplicative reduction.

David Michael Roberts (Mar 08 2024 at 00:43):

John Baez said:

I find some of this confusing (especially the part about slopes of curves being or not being in the same field as the coefficients of the curve), but this is exactly why I'd like to see examples.

In principle one could have a tangent line, defined over an algebraic closure of the original base field, that only intersects the original plane containing the elliptic curve in a single point. An analogy that occurs to me is when you have small polynomials that are irreducible over some finite finite fields but not others.

John Baez (Mar 08 2024 at 16:34):

Thanks, @Reid Barton, that really brings it down to earth in a way I understand and enjoy; it makes it easier for me to study my own examples. I hadn't really appreciated the point you and David clarified, which is in that seeking to find lines tangent to a subvariety at a point one gets polynomial equations that might not have solutions in a non-algebraically-closed field! Obvious in retrospect.

John Baez (Mar 08 2024 at 16:35):

Now I can try to start understanding an apparently harder question, which is, why do people define the $p$-local L-function of an elliptic curve in cases this way:

John Baez (Mar 08 2024 at 16:36):

1) $L_p(E,s) = 1 - a_p p^{-s} + p^{1-2s}$ if $E$ has good reduction at $p$, meaning that it's smooth as a curve defined over the field with $p$ elements.

2) $L_p(E,s) = 1 - p^{-s}$ if $E$ has 'split multiplicative reduction' at $p$.

3) $L_p(E,s) = 1 + p^{-s}$ if $E$ has 'nonsplit multiplicative reduction' at $p$

4) $L_p(E,s) = 1$ if $E$ has 'additive reduction' at $p$.

John Baez (Mar 08 2024 at 16:38):

This seems horribly baroque. The usual answer for why people do this is that they want the L-function $\prod_p L_p$ to be a modular form and you need to make these 'tweaks' for bad primes to get that to happen.

John Baez (Mar 08 2024 at 16:39):

But I feel there has to be some simpler, perhaps less easy to compute, definition of the $L$-function of an elliptic curve (or more general variety!) that doesn't involve cases.

John Baez (Mar 08 2024 at 16:41):

My bold and hopeful conjecture is that you can compute the $p$-local $L$-function knowing only the set of points of the elliptic curve over the algebraic closure $\overline{\mathbb{F}}_p$ together with the action of the Frobenius on this set.

John Baez (Mar 08 2024 at 16:42):

Someone on Mathstodon said this was true, but I was not able to squeeze out any details.

John Baez (Mar 08 2024 at 16:45):

I think I need to read some high-powered yet introductory book on $L$-functions, but I haven't found one yet. The books and papers I've seen often seem eager to dive into details, discussing various hard conjectures about $L$-functions rather than examining what $L$-functions really are (and in particular how to understand bad primes in a non-ad-hoc way).

John Baez (Mar 08 2024 at 16:48):

Being a bit of a category theorist, I naturally feel that understanding things in a simple way is important.

John Baez (Mar 08 2024 at 17:02):

If I had to do this on my own, without a book or person to save me, I'd start counting $\mathbb{F}_{p^n}$-points on some curves with bad reduction over $p$, and see how the results compare to the case of good reduction, where Hasse's theorem gives the answer:

$p^n - \alpha^n - \overline{\alpha}^n$

where $\alpha$ is a number about which a lot is known, e.g. it's an algebraic integer with $|\alpha| = p^{1/2}$.

John Baez (Mar 08 2024 at 17:03):

If I were better at Sage or something, I could do this pretty fast.

John Baez (Mar 08 2024 at 17:05):

I'm so confused that I don't even know if we get a different formula when our elliptic curve has bad reduction over $p$, but I'm hoping we do - and a different kind of formula depending on whether we have additive reduction, or split or nonsplit multiplicative reduction.

John Baez (Mar 08 2024 at 18:25):

I put out a plea for help here, which is mainly aimed at people who enjoy programming in Sage. I think someone could write a short Sage program that I could then fiddle with to count points on various curves over finite fields.

Reid Barton (Mar 08 2024 at 19:09):

This might be helpful, especially the part at the end about "Reduction types of elliptic curves":
https://ayoucis.wordpress.com/2014/11/29/classifying-one-dimensional-algebraic-groups/

If you have a plane cubic curve with a nonsmooth point P (which must be rational, see argument at the link), then any line through P meets the curve twice at P, and so it meets the curve again at one more point. That gives a mostly one-to-one correspondence between lines through P and points of the curve. The point P itself may occur 0, 1 or 2 times in this way, depending on the reduction type.

Reid Barton (Mar 08 2024 at 19:13):

I think that means that if $E$ has additive reduction, then it has $q + 1$ points over $\mathbb{F}_q$; if $E$ has split multiplicative reduction, then it has $q$ points; if $E$ has nonsplit multiplicative reduction then it has either $q$ or $q+2$ points depending on whether $q$ is an even or an odd power of $p$.

John Baez (Mar 08 2024 at 20:21):

Thanks! Wow - you completely crushed the problem! I now have a Sage program to count points on elliptic curves over finite fields, and I was just gearing up to count points in examples and discover these things myself.

John Baez (Mar 08 2024 at 20:22):

Anyway, I still plan to check out some examples, because I want to get some first-hand experience with these things.

John Baez (Mar 08 2024 at 20:23):

I will try not to remember the details of what you just said until I go through examples of each kind of curve, count points, and guess the patterns. Then I can use what you said as the "answers at back of the book".

John Baez (Mar 08 2024 at 20:25):

Anyway, it's really great that the different kinds of reduction are visible from the counts of points.

John Baez (Mar 08 2024 at 20:32):

This should make it easier to understand L-functions of elliptic curves, if all goes well.

John Baez (Mar 09 2024 at 02:22):

So something interesting happened on Mathstodon. I misread the the L-functions and modular forms database and thought

$y^2 + y = x^3$

gave a curve with a cusp when we reduce mod $p = 2$. So I counted the number of points in $\mathbb{F}_{2^n}$ for $n = 1, \dots n$ and got this table:

1 2
2 8
3 8
4 8
5 32
6 80
7 128
8 224
9 512
10 1088

John Baez (Mar 09 2024 at 02:24):

Then someone noticed that the results match this formula for the number of points:

$2^n - (\sqrt{2}\, i )^2 - (- \sqrt{2} \, i)^n$

John Baez (Mar 09 2024 at 02:25):

But this is the sort of thing Hasse's Theorem predicts when we have good reduction!

John Baez (Mar 09 2024 at 02:32):

And then I looked again and saw I had no reason to think this curve has a cusp for $p = 2$.

John Baez (Mar 09 2024 at 02:32):

I don't know if it does, but I predict it does not.

John Baez (Mar 09 2024 at 02:34):

Next I took this curve that seems to have a cusp:

$y^2 = x^3$

I haven't thought about it much over $p = 2$, but it seems to be the "walking cusp", so I imagine it has a cusp over any prime. The points should be easy to count but I'll lazily use Sage to count the points for $p = 2$:

1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
9 512
10 1024

John Baez (Mar 09 2024 at 02:36):

Obviously we're getting $2^n$ points. So I predict that when our elliptic curve gets a cusp mod $p$, i.e. the case of 'additive reduction', it has $p^n$ points over $\mathbb{F}_{p^n}$.

John Baez (Mar 09 2024 at 02:36):

At some point I'll cheat and check my guess against what Reid told me!

John Baez (Mar 09 2024 at 02:39):

Next let me take what seems like "the walking node"

$y^2 = x^2(x+1)$

.

John Baez (Mar 09 2024 at 02:41):

Without thinking at all about what really happens mod $p$ I'll count its points over $\mathbb{F}_{2^n}$:

1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
9 512
10 1024

Hmm, I'm again getting $p^n$ points!

John Baez (Mar 09 2024 at 02:42):

Maybe it doesn't really have a node when we reduce mod 2.

John Baez (Mar 09 2024 at 03:12):

Let me try mod 3:

1 4
2 8
3 28
4 80
5 244
6 728
7 2188

(It got tired and went to sleep after $3^7$.)

John Baez (Mar 09 2024 at 06:30):

I'm not sure I have the energy to figure out the pattern here... but wait: the even terms seem to be 1 less than a power of 3. The 2nd term is $3^2 - 1$, the 4th is $3^4 - 1$ and the 6th is $3^6 - 1$.

John Baez (Mar 09 2024 at 06:34):

So I'll write the nth term as $3^n$ plus a "correction term" - the kind of formula we expect from Hasse's theorem. And here are the correction terms:

John Baez (Mar 09 2024 at 06:34):

1 1
2 -1
3 1
4 -1
5 1
6 -1
7 1

John Baez (Mar 09 2024 at 06:35):

Oh, that's simple, I wonder why it didn't jump out at me. So the number of points is

$3^n - (-1)^n$

John Baez (Mar 09 2024 at 06:36):

Now this is similar to what Hasse's theorem predicts, namely

$3^n - \alpha^n - \overline{\alpha}^n$

where $\alpha$ is a complex number with $|\alpha| = \sqrt{3}$. But it's not of that form!

John Baez (Mar 09 2024 at 06:39):

If this curve were an affine line that 'crossed over itself' - like in the real picture I showed - we'd get the number of points in the affine line, $3^n$, minus $1$. But we're only getting that half the time!

John Baez (Mar 09 2024 at 06:40):

So this is weird and interesting.

John Baez (Mar 09 2024 at 06:41):

At this point I couldn't resist cheating and looking at what @Reid Barton wrote:

Reid Barton said:

if $E$ has nonsplit multiplicative reduction then it has either $q$ or $q+2$ points depending on whether $q$ is an even or an odd power of $p$.

John Baez (Mar 09 2024 at 06:43):

I swear I hadn't remembered that, Reid! But it makes me very happy since I was getting a bit freaked out. Of course your statement agrees with mine because you're counting the point at infinity while I'm not: subtracting 1 from your "$q$ if $n$ is even and $q+2$ if $n$ is odd" we get my $3^n - (-1)^n$.

John Baez (Mar 09 2024 at 06:46):

So I must indeed be getting a cusp, and one of "nonsplit multiplicative reduction" - so its two tangent lines must not be defined in the field $\mathbb{F}_3$. I could check this sometime, copying your calculation earlier.

John Baez (Mar 10 2024 at 05:00):

Reid Barton said:

if $E$ has nonsplit multiplicative reduction then it has either $q$ or $q+2$ points depending on whether $q$ is an even or an odd power of $p$.

The latter case is extremely interesting.

John Baez (Mar 10 2024 at 05:05):

First, let's see an example. Here's a cubic that has $3^n$ or $3^n+2$ points in $\mathbb{F}_{3^n}$ depending on whether $n$ is even or odd:

$y^2 = x^3 - x^2$

Apparently it's not an elliptic curve!

John Baez (Mar 10 2024 at 05:06):

But I counted the points in some finite fields of characteristic $3$:

.

John Baez (Mar 10 2024 at 05:08):

Well, here I wasn't counting the point at infinity, but if we do we see this pattern: we're getting $3^n$ points over $\mathbb{F}_{3^n}$ when $n$ is even, and $3^n+2$ points when $n$ is odd.

John Baez (Mar 10 2024 at 05:23):

For example, working projectively we get 5 points over $\mathbb{F}_3$: the solutions of $y^2 = x^3 - x^2$ are

(0,0)
(1,0)
(-1,1)
(-1,-1)

and then there's the point at infinity.

John Baez (Mar 10 2024 at 05:25):

So the blog article Reid pointed out:

• Classifying one-dimensional algebraic groups

suggests that this 5-point curve, after its node is removed, is actually a one-dimensional linear algebraic group over $\mathbb{F}_3$! So that's a group with 4 elements.

John Baez (Mar 10 2024 at 05:26):

And it's interesting, because there are two obvious one-dimensional linear algebraic groups over $\mathbb{F}_3$:

• the additive group of $\mathbb{F}_3$, which has 3 elements
• the multiplicative group of $\mathbb{F}_3$, which has 2 elements

but there are also other weird ones arising from the fact that $\mathbb{F}_3$ isn't algebraically closed. And one of these weird ones has 4 elements!

John Baez (Mar 10 2024 at 05:32):

It works like this: for some reason, for any prime power $q$ there's an epimorphism of multiplicative groups

$\mathbb{F}_{q^2}^\times \to \mathbb{F}_q^\times \to 1$

and this has a kernel so we get an exact sequence

$1 \to K \to \mathbb{F}_{q^2}^\times \to \mathbb{F}_q^\times \to 1$

and thus the cardinalities of these finite groups obey

$|\mathbb{F}_{q}^\times| = |\mathbb{F}_{q^2}^\times| \cdot |K|$

so

$|K| = \frac{q^2 - 1}{q-1} = q + 1$

John Baez (Mar 10 2024 at 05:37):

For $q = 3$ this is 4. And the part I don't understand is this: apparently the projective cubic curve coming from $y^2 = x^3 - x^2$, or at least some other cubics over $\mathbb{F}_3$, can naturally be identified with this 4-element algebraic group over $\mathbb{F}_3$.

John Baez (Mar 10 2024 at 06:38):

John Baez said:

And it's interesting, because there are two obvious one-dimensional linear algebraic groups over $\mathbb{F}_3$:

• the additive group of $\mathbb{F}_3$, which has 3 elements
• the multiplicative group of $\mathbb{F}_3$, which has 4 elements

but there are also other weird ones arising from the fact that $\mathbb{F}_3$ isn't algebraically closed. And one of these weird ones has 4 elements!

In fact that blog article claims that these are the only one-dimensional linear algebraic groups over $\mathbb{F}_3$:

• the additive group of $\mathbb{F}_3$, which has 3 elements
• the multiplicative group $\mathbb{F}_3^\times$, which has 2 elements
• the kernel of $\mathbb{F}_{3^2}^\times \to \mathbb{F}_3^\times \to 1$, which has 4 elements.

John Baez (Mar 10 2024 at 07:24):

I never explained how we get this surjective homomorphism of groups

$\mathbb{F}_{3^2}^\times \to \mathbb{F}_3^\times \to 1$

It comes from the fact that $\mathbb{F}_{3^2}$ can be constructed from $\mathbb{F}_3$ by forming numbers

$a + b i$

where $a, b \in \mathbb{F}_3$ and $i$ is a new element adjoined to this field obeying $i^2 = -1$.

John Baez (Mar 10 2024 at 07:26):

I suspect that we get a surjective homomorphism

$\mathbb{F}_{3^2}^\times \to \mathbb{F}_3^\times$

sending any element $z = a + bi$ to $z \overline{z}$, where $\overline{z} = a - bi$.

John Baez (Mar 10 2024 at 07:27):

This is a lot like the surjective homomorphism $\mathbb{C}^\times \to \mathbb{R}^\times$ sending $z$ to $z \overline{z} = |z|^2$.

John Baez (Mar 10 2024 at 07:29):

So, given this epimorphism

$\mathbb{F}_{3^2}^\times \to \mathbb{F}_3^\times$
$z \mapsto z \overline{z}$

from a 9-1 = 8 element group to a 3-1 = 2 element group, its kernel will have 8/2 = 4 elements.

John Baez (Mar 10 2024 at 16:47):

To be painfully explicit about it, since the multiplicative group of a finite field must be cyclic, we must have $\mathbb{F}_3^\times \cong \mathbb{Z}/2$ and $\mathbb{F}_9^\times \cong \mathbb{Z}/8$, so the epimorphism

$\mathbb{F}_{9}^\times \to \mathbb{F}_3^\times$

must be isomorphic to the only homomorphism from $\mathbb{Z}/8$ onto $\mathbb{Z}/2$, namely the "mod 2" map, so its kernel is $\mathbb{Z}/4$ (not the other 4-element group).

John Baez (Mar 10 2024 at 16:51):

Or to be even more painfully explicit, since $\mathbb{F}_9 \cong \mathbb{F}_3[i]/\langle i^2 + 1 \rangle$, we have

$\mathbb{F}_9^\ast = \{ \pm 1, \pm i, \pm 1 \pm i \}$

and the epimorphism $\mathbb{F}_{9}^\times \to \mathbb{F}_3^\times$ sending $z$ to $z \overline{z}$ has kernel

$\{\pm 1, \pm i\}$

where multiplication works just the same as for complex numbers.

John Baez (Mar 10 2024 at 17:20):

Putting together everything from that blog article Reid pointed out, the projective cubic over $\mathbb{F}_3$ coming from $y^2 = x^3 - x^2$ has 5 points, including a node at $x = 1$, and when we remove the node, the rest has the structure of a connected 1-dimensional linear algebraic group over $\mathbb{F}_3$, which has just 4 points, and is isomorphic as a group to $\mathbb{Z}/4$.

Reid Barton (Mar 10 2024 at 18:41):

It's good you worked out all these examples since I wasn't confident my answer was actually correct--just approximately correct.

Reid Barton (Mar 10 2024 at 18:43):

The example of $y^2 = x^3 + x^2$ over $\mathbb{F}_2$ had me worried that I overlooked some special behavior in characteristic 2--until I realized that we can rewrite the equation as $(y - x)^2 = x^3$, so it really is a cusp and not a node!

Reid Barton (Mar 10 2024 at 18:46):

By the way, you can get the other kinds of bad reduction over $\mathbb{F}_2$ with these equations:

• $y^2 + xy = x^3$
• $y^2 + xy = x^3 + x^2$

... since in characteristic 2, we can't complete the square to get rid of the $xy$ term. (These equations are chosen so that the singular point is still at $(0,0)$.)

John Baez (Mar 10 2024 at 19:19):

Nice! I actually like $\mathbb{F}_2$ better than $\mathbb{F}_3$ for a very silly reason: I'm counting points using a very crude program that simply checks all triples $(x,y,z)$ to see if they're solutions... and I'm running on a free online system with limited computing power, and $2^{3n}$ grows a lot more slowly than $3^{3n}$.

Eric M Downes (Mar 15 2024 at 20:16):

If you do decide you need to check things in a larger $\mathbb{F}_p$, I'd be happy to parallelize the brute force part and we could run these things locally. Most modern laptops have many processors. Just reach out.

John Baez (Mar 15 2024 at 23:16):

Thanks so much! @Chris Grossack (they/them) told me that Sage has a built-in command for counting points for elliptic curves (or more general curves?) over finite fields, and I bet - or at least hope - they are vastly better than anything a nonexpert on number theory algorithms would come up with.

John Baez (Mar 15 2024 at 23:19):

You can see the results of my labors so far here. It was psychologically very helpful to do some calculations even though they were crude: it made the patterns in curves over finite fields seem more concrete.

Eric M Downes (Apr 01 2024 at 10:34):

Yeah, sage would be what I would call! It is efficient, but works using only a single processor last I checked. If you wanted to check, say a finite family of curves, I would recommend using the multiprocessing library to wrap calls to Sage inside a multiprocessing pool, so that you can work on one EC per processor in parallel. This could be useful for making like a giant table you wanted to see patterns in. But it seems like you're mostly intuition-building, so probably not necessary.

John Baez (Apr 02 2024 at 18:33):

Thanks! I may eventually want to do fancier calculations with elliptic curves. Right now I've switched toward more theoretical stuff like the Tate module of an elliptic curve, and also examples related to modular curves.