### Counting Points on Elliptic Curves (Part 3)

#### Posted by John Baez

In Part 1 of this little series I showed you Wikipedia’s current definition of the $L$-function of an elliptic curve, and you were supposed to shudder in horror. In this definition the $L$-function is a product over all primes $p$. But what do we multiply in this product? There are 4 different cases, each with its own weird and unmotivated formula!

In Part 2 we studied the 4 cases. They correspond to 4 things that can happen when we look at our elliptic curve over the finite field $\mathbb{F}_{p}$: it can stay smooth, or it can become singular in 3 different ways. In each case we got a formula for number of points the resulting curve over the fields $\mathbb{F}_{p^k}$.

Now I’ll give a much better definition of the $L$-function of an elliptic curve. Using our work from last time, I’ll show that it’s equivalent to the horrible definition on Wikipedia. And eventually I may get up the nerve to improve the Wikipedia definition. Then future generations will wonder what I was complaining about.

I want to explain the $L$-function of an elliptic curve as simply as possible — thus, with a minimum of terminology and unmotivated nonsense.

The $L$-function of an elliptic curve is a slight tweak of something more fundamental: its zeta function. So we have to start there.

### The zeta function of an elliptic curve

You can define the **zeta function** of *any* gadget $S$ that assigns a finite set $S(R)$ to any finite commutative ring $R$. It goes like this:

$\zeta_S(s) = \sum_{n = 1}^\infty \frac{|Z_S(n)|}{n!} n^{-s}$

where $s$ is a complex number and the sum will converge if $Re(s)$ is big enough.

What’s $Z_S(n)$? A ring that’s a finite product of finite fields is called a **finite semisimple commutative ring**. An element of $Z_S(n)$ is a way to make the set $\{1, \dots, n\}$ into a finite semisimple commutative ring, say $R$, and choose an element of $S(R)$.

So, to define the zeta function of an elliptic curve, we just need a way for an elliptic curve $E$ to assign a finite set $E(R)$ to any finite semisimple commutative ring $R$. This is not hard. By an elliptic curve I simply mean an equation

$y^2 = P(x)$

where $P$ is a cubic equation with integer coefficients and distinct roots. When $R$ is a finite field, this equation will have a finite set of solutions in $R$, and we take those and one extra ‘point at infinity’ to be the points of our set $E(R)$. When $R$ is a general finite semsimple ring, it’s a product of finite fields, say

$R \cong F_1 \times \cdots \times F_n$

and we define

$E(R) = E(F_1) \times \cdots \times E(F_n)$

Then the **zeta function** of our elliptic curve $E$ is

$\zeta_E(s) = \sum_{n = 1}^\infty \frac{|Z_E(n)|}{n!} n^{-s}$

### The *L*-function of an elliptic curve

Later today we will calculate the zeta function of an elliptic curve. And we’ll see that it always has a special form:

$\zeta_E(s) = \frac{ \zeta(s) \zeta(s - 1)}{some \; rational \; function \; of \; s}$

where $\zeta$ is the Riemann zeta function. The denominator here is called the ** L-function** of our elliptic curve, $L(E,s)$. That’s all there is to it!

In short:

$L(E,s) = \frac{ \zeta(s) \zeta(s - 1)}{\zeta_E(s)}$

You should think of the $L$-function as the ‘interesting part’ of the zeta function of the elliptic curve — but flipped upside down, just to confuse amateurs. That’s also why we write $n^{-s}$ in the formula for the zeta function instead of $n^s$: it’s a deliberately unnatural convention designed to keep out the riff-raff.

Arbitrary conventions aside, I hope you see the $L$-function of an elliptic curve is a fairly simple thing. You might wonder why the zeta function is defined as it is, and why the zeta function of the elliptic curve has a factor of $\zeta(s) \zeta(s-1)$ in it. Those are very good questions, with good answers. But my point is this: all the *gory complexity* of the $L$-function arises when we actually try to compute it more explicitly.

Now let’s do that.

### The Euler product formula

An elliptic curve $E$ gives a finite set $E(R)$ for each finite semisimple commutative ring $R$. We need to count these sets to compute the zeta function or $L$-function of our elliptic curve. But we have set things up so that

$E(R \times R') \cong E(R) \times E(R')$

Since every finite semisimple commutative ring is a product of finite fields, this lets us focus on counting $E(R)$ when $R$ is a finite field. And since every finite field has a prime power number of elements, we can tackle this counting problem ‘one prime at a time’.

If we carry this through, we get an interesting formula for the zeta function of an elliptic curve. In fact it’s a very general thing:

**Euler Product Formula.** Suppose $S$ is any functor from finite commutative rings to finite sets such that $S(R \times R') \cong S(R) \times S(R')$. Then

$\zeta_S(s) = \prod_p \exp \left( \sum_{k = 1}^\infty \frac{|S(\mathbb{F}_{p^k})|}{k} p^{-k s} \right)$

where we take the product over all primes $p$, and $\mathbb{F}_{p^k}$ is the field with $p^k$ elements.

I wrote up a proof here:

so check it out if you want. I was not trying to make the argument look as simple as possible, but it’s really quite easy given what I’ve said: you can probably work it out yourself.

So: the zeta function of an elliptic curve $E$ is a product over primes. The factor for the prime $p$ is called the **local zeta function**

$Z_p(E,s) = \exp \left( \sum_{k = 1}^\infty \frac{|E(\mathbb{F}_{p^k})|}{k} p^{-k s} \right)$

To compute this, we need to know the numbers $|E(\mathbb{F}_{p^k})|$. Luckily we worked these out last time! But there are four cases.

In every case we have

$|E(\mathbb{F}_{p^k})| = p^k + 1 + c(p,k)$

where $c(p,k)$ is some sort of ‘correction’. If the correction $c(p,k)$ is zero, we get

$\begin{array}{ccl} Z_p(E,s) &=& \displaystyle{ \exp \left(\sum_{k = 1}^\infty \frac{p^k + 1}{k} p^{-k s} \right) } \\ \\ &=& \displaystyle{ \exp \left( -\ln(1 - p^{-s + 1}) - \ln(1 - p^{-s}) \right) } \\ \\ &=& \displaystyle{ \frac{1}{(1 - p^{-s + 1})(1 - p^{-s}) } } \end{array}$

I did the sum pretty fast, but not because I’m good at sums — merely to keep you from getting bored. To do it yourself, all you need to know is the Taylor series for the logarithm.

To get the zeta function of our elliptic curve we multiply all the local zeta functions $Z_p(E,s)$. So if *all* the corrections $c(p,k)$ were zero, we’d get

$Z(E,s) = \prod_p \frac{1}{1 - p^{-s + 1}} \prod_p \frac{1}{1 - p^{-s} } = \zeta(s-1) \zeta(s)$

Here I used the Euler product formula for the Riemann zeta function.

This is precisely why folks define the $L$-function of an elliptic curve to be

$L(E,s)^{-1} = \frac{\zeta_E(s)}{ \zeta(s) \zeta(s - 1)}$

It lets us focus on the effect of the corrections!. Well, it doesn’t explain that stupid reciprocal on the left-hand side, which is just a convention — but apart from that, we’re taking the zeta function of the elliptic curve and dividing out by what we’d get if all the corrections $c(p,k)$ were zero. So, if you think about it a bit, we have

$L(E,s)^{-1} = \prod_p \exp \left( \sum_{k = 1}^\infty \frac{c(p,k)}{k} p^{-k s} \right)$

It’s like the Euler product formula for the zeta function, but using only the corrections $c(p,k)$ instead of the full count of points $|E(\mathbb{F}_{p^k})|$.

As you can see, the $L$-function is a product of **local L-functions**

$L_p(E,s)^{-1} = \exp \left( \sum_{k = 1}^\infty \frac{c(p,k)}{k} p^{-k s}\right)$

So let’s work those out! There are four cases.

### The local zeta function of an elliptic curve: additive reduction

If our elliptic curve gets a cusp over $\mathbb{F}_p$, we say it has **additive reduction**. In this case we saw in Theorem 2 last time that

$|E(\mathbb{F}_{p^k})| = p^k + 1$

So in this case the correction vanishes:

$c(p,k) = 0$

This makes the local $L$-function very simple:

$L_p(E,s)^{-1} = \exp \left( \sum_{k = 1}^\infty \frac{c(p,k)}{k} p^{-k s}\right) = 1$

### The local zeta function of an elliptic curve: split multiplicative reduction

If our elliptic curve gets a node over $\mathbb{F}_p$ and the two lines tangent to this node have slopes defined in $\mathbb{F}_p$, we say our curve has **split multiplicative reduction**.
In this case we saw in Theorem 3 last time that

$|E(\mathbb{F}_{p^k})| = p^k$

So in this case, the correction is $-1$:

$c(p,k) = -1$

This gives

$\begin{array}{ccl} L_p(E,s)^{-1} &=& \displaystyle{ \exp \left( \sum_{k = 1}^\infty \frac{1}{k} p^{-k s}\right) } \\ \\ &=& \displaystyle{ \exp \left( \ln(1 - p^{-s}) \right) } \\ \\ &=& 1 - p^{-s} \end{array}$

Again I used my profound mastery of Taylor series of the logarithm to do the sum.

### The local zeta function of an elliptic curve: split multiplicative reduction

If our elliptic curve gets a node over $\mathbb{F}_p$ and the two lines tangent to this node have slopes that are not defined in $\mathbb{F}_p$, we say our curve has **nonsplit multiplicative reduction**. In this case we saw in Theorem 4 last time that

$|E(\mathbb{F}_{p^k})| = p^k + 1 - (-1)^k$

In this case the correction is more interesting:

$c(p,k) = -(-1)^k$

This gives

$\begin{array}{ccl} L_p(E,s)^{-1} &=& \displaystyle{ \exp \left( -\sum_{k = 1}^\infty \frac{(-1)^k}{k} p^{-k s}\right) } \\ \\ &=& \displaystyle{ \exp \left( \ln(1 + p^{-s}) \right) } \\ \\ &=& 1 + p^{-s} \end{array}$

Again, I just used the Taylor series of the log function.

## The local zeta function of an elliptic curve: good reduction

If our elliptic curve stays smooth over $\mathbb{F}_p$, we say it has **good reduction.** Ironically this gives the most complicated local zeta function. In Theorem 1 last time we saw

$|E(\mathbb{F}_{p^k})| = p^k - \alpha^k - \overline{\alpha}^k + 1$

where $\alpha$ is a complex number with $\alpha \overline{\alpha} = p$. We didn’t prove this, we literally just *saw* it: it’s a fairly substantial result due to Hasse.

So, in this case the correction is

$c(p,k) = \alpha^k + \overline{\alpha}^k$

This gives

$\begin{array}{ccl} L_p(E,s)^{-1} &=& \displaystyle{ \exp \left( -\sum_{k = 1}^\infty \frac{\alpha^k + \overline{\alpha}^k}{k} p^{-k s}\right) } \\ \\ &=& \displaystyle{ \exp \left( \ln\left(1 - \alpha p^{-s}\right) \; + \; \ln\left(1 - \overline{\alpha} p^{-s}\right) \right) } \\ \\ &=& (1 - \alpha p^{-s})(1 - \overline{\alpha} p^{-s}) \end{array}$

Again I just used the Taylor series of the log function. I’m sure glad I went to class that day.

But we can get a bit further using $\alpha \overline{\alpha} = p$:

$\begin{array}{ccl} L_p(E,s)^{-1} &=& (1 - \alpha p^{-s})(1 - \overline{\alpha} p^{-s}) \\ &=& 1 - (\alpha + \overline{\alpha})p^{-s} + p^{1-2s} \end{array}$

At this point people usually notice that

$|E(\mathbb{F}_{p})| = p - \alpha - \overline{\alpha} + 1$

so

$\alpha + \overline{\alpha} = p + 1 - |E(\mathbb{F}_{p})|$

Thus, you can compute this number using just the number of points of our curve over $\mathbb{F}_p$. And to be cute, people call this number something like $a_p(E)$. So in the end, for elliptic curves of good reduction over the prime $p$ we have

$L_p(E,s)^{-1} = 1 - a_p(E) p^{-s} + p^{1-2s}$

Whew, we’re done!

### The *L*-function of an elliptic curve, revisited

Okay, now we can summarize all our work in an *explicit formula* for the $L$-function of an elliptic curve.

**Theorem.** The $L$-function of an elliptic curve $E$ equals

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

where:

1) $L_p(E,s) = 1 - a_p(E) p^{-s} + p^{1-2s}$ if $E$ remains smooth over $\mathbb{F}_p$. Here $a_p(E)$ is $p + 1$ minus the number of points of $E$ over $\mathbb{F}_p$.

2) $L_p(E,s) = 1$ if $E$ gets a cusp over $\mathbb{F}_p$.

3) $L_p(E,s) = 1 - p^{-s}$ if $E$ gets a node over $\mathbb{F}_p$, and the two tangent lines to this node have slopes that are defined in $\mathbb{F}_p$.

4) $L_p(E,s) = 1 + p^{-s}$ if $E$ we gets a node over $\mathbb{F}_p$, but the two tangent lines to this node have slopes that are not defined in $\mathbb{F}_p$.

My god! This is exactly what I showed you in Part 1. So this rather elaborate theorem is what some people run around calling the *definition* of the $L$-function of an elliptic curve!

## Re: Counting Points on Elliptic Curves (Part 3)

Is this really called a “simple ring”? Wouldn’t semisimple be a better name?

Great stuff by the way.