# A parlor trick for SET

Tai-Danae Bradley is one of the hosts of PBS Infinite Series, a delightful series of vignettes into fun parts of math. The video below is about the same of SET, a favorite among mathematicians. Specifically, Tai-Danae explains how SET cards lie in (using more technical jargon) a vector space over a finite field, and that valid sets correspond to lines. If you don’t immediately know how this would work, watch the video.

In this post I want to share a parlor trick for SET that I originally heard from Charlotte Chan. It uses the same ideas from the video above, which I’ll only review briefly.

In the game of SET you see a board of cards like the following, and players look for sets.

Image source: theboardgamefamily.com

A valid set is a triple of cards where, feature by feature, the characteristics on the cards are either all the same or all different. A valid set above is {one empty blue oval, two solid blue ovals, three shaded blue ovals}. The feature of “fill” is different on all the cards, but the feature of “color” is the same, etc.

In a game of SET, the cards are dealt in order from a shuffled deck, players race to claim sets, removing the set if it’s valid, and three cards are dealt to replace the removed set. Eventually the deck is exhausted and the game is over, and the winner is the player who collected the most sets.

There are a handful of mathematical tricks you can use to help you search for sets faster, but the parlor trick in this post adds a fun variant to the end of the game.

Play the game of SET normally, but when you get down to the last card in the deck, don’t reveal it. Keep searching for sets until everyone agrees no visible sets are left. Then you start the variant: the first player to guess the last un-dealt card in the deck gets a bonus set.

The math comes in when you discover that you don’t need to guess, or remember anything about the game that was just played! A clever stranger could walk into the room at the end of the game and win the bonus point.

Theorem: As long as every player claimed a valid set throughout the game, the information on the remaining board uniquely determines the last (un-dealt) card.

Before we get to the proof, some reminders. Recall that there are four features on a SET card, each of which has three options. Enumerate the options for each feature (e.g., {Squiggle, Oval, Diamond} = {0, 1, 2}).

While we will not need the geometry induced by this, this implies each card is a vector in the vector space $\mathbb{F}_3^4$, where $\mathbb{F}_3 = \mathbb{Z}/3\mathbb{Z}$ is the finite field of three elements, and the exponent means “dimension 4.” As Tai-Danae points out in the video, each SET is an affine line in this vector space. For example, if this is the enumeration:

Source: “The Joy of Set

Then using the enumeration, a set might be given by

$\displaystyle \{ (1, 1, 1, 1), (1, 2, 0, 1), (1, 0, 2, 1) \}$

The crucial feature for us is that the vector-sum (using the modular field arithmetic on each entry) of the cards in a valid set is the zero vector $(0, 0, 0, 0)$. This is because $1+1+1 = 0, 2+2+2 = 0,$ and $1+2+3=0$ are all true mod 3.

Proof of Theorem. Consider the vector-valued invariant $S_t$ equal to the sum of the remaining cards after $t$ sets have been taken. At the beginning of the game the deck has 81 cards that can be partitioned into valid sets. Because each valid set sums to the zero vector, $S_0 = (0, 0, 0, 0)$. Removing a valid set via normal play does not affect the invariant, because you’re subtracting a set of vectors whose sum is zero. So $S_t = 0$ for all $t$.

At the end of the game, the invariant still holds even if there are no valid sets left to claim. Let $x$ be the vector corresponding to the last un-dealt card, and $c_1, \dots, c_n$ be the remaining visible cards. Then $x + \sum_{i=1}^n c_i = (0,0,0,0)$, meaning $x = -\sum_{i=1}^n c_i$.

$\square$

I would provide an example, but I want to encourage everyone to play a game of SET and try it out live!

Charlotte, who originally showed me this trick, was quick enough to compute this sum in her head. So were the other math students we played SET with. It’s a bit easier than it seems since you can do the sum feature by feature. Even though I’ve known about this trick for years, I still require a piece of paper and a few minutes.

Because this is Math Intersect Programming, the reader is encouraged to implement this scheme as an exercise, and simulate a game of SET by removing randomly chosen valid sets to verify experimentally that this scheme works.

Until next time!

# The Welch-Berlekamp Algorithm for Correcting Errors in Data

In this post we’ll implement Reed-Solomon error-correcting codes and use them to play with codes. In our last post we defined Reed-Solomon codes rigorously, but in this post we’ll focus on intuition and code. As usual the code and data used in this post is available on this blog’s Github page.

The main intuition behind Reed-Solomon codes (and basically all the historically major codes) is

Error correction is about adding redundancy, and polynomials are a really efficient way to do that.

Here’s an example of what we’ll do in the post. Say you have a space probe flying past Mars taking photographs like this one

Courtesy of NASA’s Viking Orbiter.

Unfortunately you know that if you send the images back to Earth via radio waves, the signal will get corrupted by cosmic something-or-other and you’ll end up with an image like this.

How can you recover from errors like this? You could do something like repeat each pixel twice in the message so that if one is corrupted the other will get through. But still, every now and then both pixels in a row will be corrupted and it’s twice as inefficient.

The idea of error-correcting codes is to find a way to encode a message so that it adds a lot of redundancy without adding too much extra information to the message. The name of the game is to optimize the tradeoff between how much redundancy you get and how much longer the message needs to be, while still being able to efficiently decode the encoded message.

A solid technique turns out to be: use polynomials. Even though you’d think polynomials are too simple (we teach them starting in the 7th grade these days!) they turn out to have remarkable properties. The most important of which is:

if you give me a bunch of points in the plane with different $x$ coordinates, they uniquely define a polynomial of a certain degree.

This fact is called polynomial interpolation. We used it in a previous post to share secrets, if you’re interested.

What makes polynomials great for error correction is that you can take a fixed polynomial (think, the message) and “encode” it as a list of points on that polynomial. If you include enough, then you can get back the original polynomial from the points alone. And the best part, for each two additional points you include above the minimum, you get resilience to one additional error no matter where it happens in the message. Another way to say this is, even if some of the points in your encoded message are wrong (the numbers are modified by an adversary or random noise), as long as there aren’t too many errors there is an algorithm that can recover the errors.

That’s what makes polynomials so much better than the naive idea of repeating every pixel twice: once you allow for three errors you run the risk of losing a pixel, but you had to double your communication costs. With a polynomial-based approach you’d only need to store around six extra pixels worth of data to get resilience to three errors that can happen anywhere. What a bargain!

Here’s the official theorem about Reed-Solomon codes:

Theorem: There is an efficient algorithm which, when given points $(a_1, b_1), \dots, (a_n, b_n)$ with distinct $a_i$ has the following property. If there is a polynomial of degree $d$ that passes through at least $n/2 + d/2$ of the given points, then the algorithm will output the polynomial.

So let’s implement the encoder, decoder, and turn the theorem into code!

## Implementing the encoder

The way you write a message of length $k$ as a polynomial is easy. Pick a large prime integer $p$ and from now on we’ll do all our arithmetic modulo $p$. Then encode each character $c_0, c_1, \dots, c_{k-1}$ in the message as an integer between 0 and $p-1$ (this is why $p$ needs to be large enough), and the polynomial representing the message is

$m(x) = c_0 + c_1x + c_2x^2 + \dots + c_{k-1}x^{k-1}$

If the message has length $k$ then the polynomial will have degree $k-1$.

Now to encode the message we just pick a bunch of $x$ values and plug them into the polynomial, and record the (input, output) pairs as the encoded message. If we want to make things simple we can just require that you always pick the $x$ values $0, 1, \dots, n$ for some choice of $n \leq p$.

A quick skippable side-note: we need $p$ to be prime so that our arithmetic happens in a field. Otherwise, we won’t necessarily get unique decoded messages.

Back when we discussed elliptic curve cryptography (ironically sharing an acronym with error correcting codes), we actually wrote a little library that lets us seamlessly represent polynomials with “modular arithmetic coefficients” in Python, which in math jargon is a “finite field.” Rather than reinvent the wheel we’ll just use that code as a black box (full source in the Github repo). Here are some examples of using it.

>>> from finitefield.finitefield import FiniteField
>>> F13 = FiniteField(p=13)
>>> a = F13(7)
>>> a+9
3 (mod 13)
>>> a*a
10 (mod 13)
>>> 1/a
2 (mod 13)


A programming aside: once you construct an instance of your finite field, all arithmetic operations involving instances of that type will automatically lift integers to the appropriate type. Now to make some polynomials:

>>> from finitefield.polynomial import polynomialsOver
>>> F = FiniteField(p=13)
>>> P = polynomialsOver(F)
>>> g = P([1,3,5])
>>> g
1 + 3 t^1 + 5 t^2
>>> g*g
1 + 6 t^1 + 6 t^2 + 4 t^3 + 12 t^4
>>> g(100)
4 (mod 13)


Now to fix an encoding/decoding scheme we’ll call $k$ the size of the unencoded message, $n$ the size of the encoded message, and $p$ the modulus, and we’ll fix these programmatically when the encoder and decoder are defined so we don’t have to keep carrying these data around.

def makeEncoderDecoder(n, k, p):
Fp = FiniteField(p)
Poly = polynomialsOver(Fp)

def encode(message):
...

def decode(encodedMessage):
...

return encode, decode


Encode is the easier of the two.

def encode(message):
thePoly = Poly(message)
return [(Fp(i), thePoly(Fp(i))) for i in range(n)]


Technically we could remove the leading Fp(i) from each tuple, since the decoder algorithm can assume we’re using the first $n$ integers in order. But we’ll leave it in and define the decode function more generically.

After we define how the decoder should work in theory we’ll run through a simple example step by step. Now on to the decoder.

## The decoding algorithm, Berlekamp-Welch

There are a lot of different decoding algorithms for various error correcting codes. The one we’ll implement is called the Berlekamp-Welch algorithm, but before we get to it we should mention a much simpler algorithm that will work when there are only a few errors.

To remind us of notation, call $k$ the length of the message, so that $k-1$ is the degree of the polynomial we used to encode it. And $n$ is the number of points we used in the encoding. Call the encoded message $M$ as it’s received (as a list of points, possibly with errors).

In the simple method what you do is just randomly pick $k$ points from $M$, do polynomial interpolation on the chosen points to get some polynomial $g$, and see if $g$ agrees with most of the points in $M$. If there really are few errors, then there’s a good chance the randomly chosen points won’t have any errors in them and you’ll win. If you get unlucky and pick some points with errors, then the $g$ you get won’t agree with most of $M$ and you can throw it out and try again. If you get really unlucky and a bad $g$ does agree with most of $M$, then you just run this procedure a few hundred times and take the $g$ you get most often. But again, this only works with a small number of errors and while it could be good enough for many applications, don’t bet your first-born child’s life on it working. Or even your favorite pencil, for that matter. We’re going to implement Berlekamp-Welch so you can win someone else’s favorite pencil. You’re welcome.

Exercise: Implement the simple decoding algorithm and test it on some data.

Suppose we are guaranteed that there are exactly $e < \frac{n-k+1}{2}$ errors in our received message $M = (a_1, b_1, \dots, a_n, b_n)$. Call the polynomial that represents the original message $P$. In other words, we have that $P(a_i) = b_i$ for all but $e$ of the points in $M$.

There are two key ingredients in the algorithm. The first is called the error locator polynomial. We’ll call this polynomial $E(x)$, and it’s just defined by being zero wherever the errors occurred. In symbols, $E(a_i) = 0$ whenever $P(a_i) \neq b_i$. If we knew where the errors occurred, we could write out $E(x)$ explicitly as a product of terms like $(x-a_i)$. And if we knew $E$ we’d also be done, because it would tell us where the errors were and we could do interpolation on all the non-error points in $M$.

So we’re going to have to study $E$ indirectly and use it to get $P$. One nice property of $E(x)$ is the following

$\displaystyle b_i E(a_i) = P(a_i)E(a_i),$

which is true for every pair $(a_i, b_i) \in M$. Indeed, by definition when $P(a_i) \neq b_i$ then $E(a_i) = 0$ so both sides are zero. Now we can use a technique called linearization. It goes like this. The product $P(x) E(x)$, i.e. the right-hand-side of the above equation, is a polynomial, say $Q(x)$, of larger degree ($e + k - 1$). We get the equation for all $i$:

$\displaystyle b_i E(a_i) = Q(a_i)$

Now $E$, $Q$, and $P$ are all unknown, but it turns out that we can actually find $E$ and $Q$ efficiently. Or rather, we can’t guarantee we’ll find $E$ and $Q$ exactly, instead we’ll find two polynomials that have the same quotient as $Q(x)/E(x) = P(x)$. Here’s how that works.

Say we wrote out $E(x)$ as a generic polynomial of degree $e$ and $Q(x)$ as a generic polynomial of degree $e+k-1$. So their coefficients are unspecified variables. Now we can plug in all the points $a_i, b_i$ to the equations $b_i E(a_i) = Q(a_i)$, and this will form a linear system of $2e + k-1$ unknowns ($e$ unknowns come from $E(x)$ and $e+k-1$ come from $Q(x)$).

Now we know that this system has good solution, because if we take the true error locator polynomial and $Q = E(x)P(x)$ with the true $P(x)$ we win. The worry is that we’ll solve this system and get two different polynomials $Q', E'$ whose quotient will be something crazy and unrelated to $P$. But as it turns out this will never happen, and any solution will give the quotient $P$. Here’s a proof you can skip if you hate proofs.

Proof. Say you have two pairs of solutions to the system, $(Q_1, E_1)$ and $(Q_2, E_2)$, and you want to show that $Q_1/E_1 = Q_2/E_2$. Well, they might not be divisible, but we can multiply the previous equation through to get $Q_1E_2 = Q_2E_1$. Now we show two polynomials are equal in the same way as always: subtract and show there are too many roots. Define $R(x) = Q_1E_2 - Q_2E_1$. The claim is that $R(x)$ has $n$ roots, one for every point $(a_i, b_i)$. Indeed,

$\displaystyle R(a_i) = (b_i E_1(a_i))E_2(a_i) - (b_iE_2(a_i)) E_1(a_i) = 0$

But the degree of $R(x)$ is $2e + k - 1$ which is less than $n$ by the assumption that $e < \frac{n-k+1}{2}$. So $R(x)$ has too many roots and must be the zero polynomial, and the two quotients are equal.

$\square$

So the core python routine is just two steps: solve the linear equation, and then divide two polynomials. However, it turns out that no python module has any decent support for solving linear systems of equations over finite fields.  Luckily, I wrote a linear solver way back when and so we’ll adapt it to our purposes. I’ll leave out the gory details of the solver itself, but you can see them in the source for this post. Here is the code that sets up the system

   def solveSystem(encodedMessage):
for e in range(maxE, 0, -1):
ENumVars = e+1
QNumVars = e+k
def row(i, a, b):
return ([b * a**j for j in range(ENumVars)] +
[-1 * a**j for j in range(QNumVars)] +
[0]) # the "extended" part of the linear system

system = ([row(i, a, b) for (i, (a,b)) in enumerate(encodedMessage)] +
[[0] * (ENumVars-1) + [1] + [0] * (QNumVars) + [1]])
# ensure coefficient of x^e in E(x) is 1

solution = someSolution(system, freeVariableValue=1)
E = Poly([solution[j] for j in range(e + 1)])
Q = Poly([solution[j] for j in range(e + 1, len(solution))])

P, remainder = Q.__divmod__(E)
if remainder == 0:
return Q, E

raise Exception("found no divisors!")

def decode(encodedMessage):
Q,E = solveSystem(encodedMessage)

P, remainder = Q.__divmod__(E)
if remainder != 0:
raise Exception("Q is not divisibly by E!")

return P.coefficients


## A simple example

Now let’s go through an extended example with small numbers. Let’s work modulo 7 and say that our message is

2, 3, 2 (mod 7)


In particular, $k=3$ is the length of the message. We’ll encode it as a polynomial in the way we described:

$\displaystyle m(x) = 2 + 3x + 2x^2 (\mod 7)$

If we pick $n = 5$, then we will encode the message as a sequence of five points on $m(x)$, namely $m(0)$ through $m(4)$.

[[0, 2], [1, 0], [2, 2], [3, 1], [4, 4]] (mod 7)


Now let’s add a single error. First remember that our theoretical guarantee says that we can correct any number of errors up to $\frac{n+k-1}{2} - 1$, which in this case is $[(5+3-1) / 2] - 1 = 2$, so we can definitely correct one error. We’ll add 1 to the third point, giving the received corrupted message as

[[0, 2], [1, 0], [2, 3], [3, 1], [4, 4]] (mod 7)


Now we set up the system of equations $b_i E(a_i) = Q(a_i)$ for all $(a_i, b_i)$ above. Rewriting the equations as $b_iE(a_i) - Q(a_i) = 0$, and adding as the last equation the constraint that the coefficient of $x^e$ is $1$, so that we get a “generic” error locator polynomial of the right degree. The columns represent the variables, with the last column being the right-hand-side of the equality as is the standard for Gaussian elimination.

# e0 e1 q0 q1 q2 q3
[
[2, 0, 6, 0, 0, 0, 0],
[0, 0, 6, 6, 6, 6, 0],
[3, 6, 6, 5, 3, 6, 0],
[1, 3, 6, 4, 5, 1, 0],
[4, 2, 6, 3, 5, 6, 0],
[0, 1, 0, 0, 0, 0, 1],
]


Then we do row-reduction to get

[
[1, 0, 0, 0, 0, 0, 5],
[0, 1, 0, 0, 0, 0, 1],
[0, 0, 1, 0, 0, 0, 3],
[0, 0, 0, 1, 0, 0, 3],
[0, 0, 0, 0, 1, 0, 6],
[0, 0, 0, 0, 0, 1, 2]
]


And reading off the solution gives $E(x) = 5 + x$ and $Q(x) = 3 + 3x + 6x^2 + 2x^3$. Note in particular that the $E(x)$ given in this solution is the true error locator polynomial, but it is not guaranteed to be so! Either way, the quotient of the two polynomials is exactly $m(x) = 2 + 3x + 2x^2$ which gives back the original message.

There is one catch here: how does one determine the value of $e$ to use in setting up the system of linear equations? It turns out that an upper bound on $e$ will work just fine, so long as the upper bound you use agrees with the theoretical maximum number of errors allowed (see the Singleton bound from last time). The effect of doing this is that the linear system ends up with some number of free variables that you can set to arbitrary values, and these will correspond to additional shared roots of $E(x)$ and $Q(x)$ that cancel out upon dividing.

## A larger example

Now it’s time for a sad fact. I tried running Welch-Berlekamp on an encoded version of the following tiny image:

And it didn’t finish after running all night.

Berlekamp-Welch is a slow algorithm for decoding Reed-Solomon codes because it requires one to solve a large system of equations. There’s at least one equation for each pixel in a black and white image! To get around this one typically encodes blocks of pixels together into one message character (since $p$ is larger than $n > k$ there is lots of space), and apparently one can balance it to minimize the number of equations. And finally, a nontrivial inefficiency comes from our implementation of everything in Python without optimizations. If we rewrote everything in C++ or Go and fixed the prime modulus, we would likely see reasonable running times. There are also asymptotically much faster methods based on the fast Fourier transform, and in the future we’ll try implementing some of these. For the dedicated reader, these are all good follow-up projects.

For now we’ll just demonstrate that it works by running it on a larger sample of text, the introductory paragraphs of To Kill a Mockingbird:

def tkamTest():
message = '''When he was nearly thirteen, my brother Jem got his arm badly broken at the elbow.  When it healed, and Jem's fears of never being able to play football were assuaged, he was seldom   self-conscious about his injury. His left arm was somewhat shorter than his right; when he stood or walked, the back of his hand was at right angles to his body, his thumb parallel to his thigh. He   couldn't have cared less, so long as he could pass and punt.'''

k = len(message)
n = len(message) * 2
p = 2087
integerMessage = [ord(x) for x in message]

enc, dec, solveSystem = makeEncoderDecoder(n, k, p)
print("encoding...")
encoded = enc(integerMessage)

e = int(k/2)
print("corrupting...")
corrupted = corrupt(encoded[:], e, 0, p)

print("decoding...")
Q,E = solveSystem(corrupted)
P, remainder = (Q.__divmod__(E))

recovered = ''.join([chr(x) for x in P.coefficients])
print(recovered)


Running this with unix time produces the following:

encoding...
corrupting...
decoding...
When he was nearly thirteen, my brother Jem got his arm badly broken at the elbow. When it healed, and Jem's fears of never being able to play football were assuaged, he was seldom self-conscious about his injury. His left arm was somewhat shorter than his right; when he stood or walked, the back of his hand was at right angles to his body, his thumb parallel to his thigh. He couldn't have cared less, so long as he could pass and punt.

real	82m9.813s
user	81m18.891s
sys	0m27.404s


So it finishes in “only” an hour or so.

In any case, the decoding algorithm is an interesting one. In future posts we’ll explore more efficient algorithms and faster implementations.

Until then!

Posts in this series:

# Programming with Finite Fields

Back when I was first exposed to programming language design, I decided it would be really cool if there were a language that let you define your own number types and then do all your programming within those number types. And since I get excited about math, I think of really exotic number types (Boolean rings, Gaussian integers, Octonions, oh my!). I imagined it would be a language feature, so I could do something like this:

use gaussianintegers as numbers

x = 1 + i
y = 2 - 3i
print(x*y)

z = 2 + 3.5i     # error


I’m not sure why I thought this would be so cool. Perhaps I felt like I would be teaching a computer math. Or maybe the next level of abstraction in playing god by writing programs is to play god by designing languages (and I secretly satisfy a massive god complex by dictating the actions of my computer).

But despite not writing a language of my own, programming with weird number systems still has a special place in my heart. It just so happens that we’re in the middle of a long series on elliptic curves, and in the next post we’ll actually implement elliptic curve arithmetic over a special kind of number type (the finite field). In this post, we’ll lay the groundwork by implementing number types in Python that allow us to work over any finite field. This is actually a pretty convoluted journey, and to be totally rigorous we’d need to prove a bunch of lemmas, develop a bunch of ring theory, and prove the correctness of a few algorithms involving polynomials.

Instead of taking the long and winding road, we’ll just state the important facts with links to proofs, prove the easy stuff, and focus more heavily than usual on the particular Python implementation details. As usual, all of the code used in this post is available on this blog’s Github page.

## Integers Modulo Primes

The simples kind of finite field is the set of integers modulo a prime. We’ve dealt with this number field extensively on this blog (in groups, rings, fields, with RSA, etc.), but let’s recall what it is. The modulo operator $\mod$ (in programming it’s often denoted %) is a binary operation on integers such that $x \mod y$ is the unique positive remainder of $x$ when divided by $y$.

Definition: Let $p$ be a prime number. The set $\mathbb{Z}/p$ consists of the numbers $\left \{ 0, 1, \dots, p-1 \right \}$. If you endow it with the operations of addition (mod $p$) and multiplication (mod $p$), it forms a field.

To say it’s a field is just to say that arithmetic more or less behaves the way we expect it to, and in particular that every nonzero element has a (unique) multiplicative inverse. Making a number type for $\mathbb{Z}/p$ in Python is quite simple.

def IntegersModP(p):
class IntegerModP(FieldElement):
def __init__(self, n):
self.n = n % p
self.field = IntegerModP

def __add__(self, other): return IntegerModP(self.n + other.n)
def __sub__(self, other): return IntegerModP(self.n - other.n)
def __mul__(self, other): return IntegerModP(self.n * other.n)
def __truediv__(self, other): return self * other.inverse()
def __div__(self, other): return self * other.inverse()
def __neg__(self): return IntegerModP(-self.n)
def __eq__(self, other): return isinstance(other, IntegerModP) and self.n == other.n
def __abs__(self): return abs(self.n)
def __str__(self): return str(self.n)
def __repr__(self): return '%d (mod %d)' % (self.n, self.p)

def __divmod__(self, divisor):
q,r = divmod(self.n, divisor.n)
return (IntegerModP(q), IntegerModP(r))

def inverse(self):
...?

IntegerModP.p = p
IntegerModP.__name__ = 'Z/%d' % (p)
return IntegerModP


We’ve done a couple of things worth note here. First, all of the double-underscore methods are operator overloads, so they are called when one tries to, e.g., add two instances of this class together. We’ve also implemented a division algorithm via __divmod__ which computes a (quotient, remainder) pair for the appropriate division. The built in Python function divmod function does this for integers, and we can overload it for a custom type. We’ll write a more complicated division algorithm later in this post. Finally, we’re dynamically creating our class so that different primes will correspond to different types. We’ll come back to why this encapsulation is a good idea later, but it’s crucial to make our next few functions reusable and elegant.

Here’s an example of the class in use:

>>> mod7 = IntegersModP(7)
>>> mod7(3) + mod7(6)
2 (mod 7)


The last (undefined) function in the IntegersModP class, the inverse function, is our only mathematical hurdle. Luckily, we can compute inverses in a generic way, using an algorithm called the extended Euclidean algorithm. Here’s the mathematics.

Definition: An element $d$ is called a greatest common divisor (gcd) of $a,b$ if it divides both $a$ and $b$, and for every other $z$ dividing both $a$ and $b$, $z$ divides $d$. For $\mathbb{Z}/p$ gcd’s and we denote it as $\gcd(a,b)$. [1]

Note that we called it ‘a’ greatest common divisor. In general gcd’s need not be unique, though for integers one often picks the positive gcd. We’ll actually see this cause a tiny programmatic bug later in this post, but let’s push on for now.

Theorem: For any two integers $a,b \in \mathbb{Z}$ there exist unique $x,y \in \mathbb{Z}$ such that $ax + by = \gcd(a,b)$.

We could beat around the bush and try to prove these things in various ways, but when it comes down to it there’s one algorithm of central importance that both computes the gcd and produces the needed linear combination $x,y$. The algorithm is called the Euclidean algorithm. Here is a simple version that just gives the gcd.

def gcd(a, b):
if abs(a) < abs(b):
return gcd(b, a)

while abs(b) > 0:
q,r = divmod(a,b)
a,b = b,r

return a


This works by the simple observation that $\gcd(a, aq+r) = \gcd(a,r)$ (this is an easy exercise to prove directly). So the Euclidean algorithm just keeps applying this rule over and over again: take the remainder when dividing the bigger argument by the smaller argument until the remainder becomes zero. Then the $\gcd(x,0) = x$ because everything divides zero.

Now the so-called ‘extended’ Euclidean algorithm just keeps track of some additional data as it goes (the partial quotients and remainders). Here’s the algorithm.

def extendedEuclideanAlgorithm(a, b):
if abs(b) > abs(a):
(x,y,d) = extendedEuclideanAlgorithm(b, a)
return (y,x,d)

if abs(b) == 0:
return (1, 0, a)

x1, x2, y1, y2 = 0, 1, 1, 0
while abs(b) > 0:
q, r = divmod(a,b)
x = x2 - q*x1
y = y2 - q*y1
a, b, x2, x1, y2, y1 = b, r, x1, x, y1, y

return (x2, y2, a)


Indeed, the reader who hasn’t seen this stuff before is encouraged to trace out a run for the numbers 4864, 3458. Their gcd is 38 and the two integers are 32 and -45, respectively.

How does this help us compute inverses? Well, if we want to find the inverse of $a$ modulo $p$, we know that their gcd is 1. So compute the $x,y$ such that $ax + py = 1$, and then reduce both sides mod $p$. You get $ax + 0 = 1 \mod p$, which means that $x \mod p$ is the inverse of $a$. So once we have the extended Euclidean algorithm our inverse function is trivial to write!

def inverse(self):
x,y,d = extendedEuclideanAlgorithm(self.n, self.p)
return IntegerModP(x)


And indeed it works as expected:

>>> mod23 = IntegersModP(23)
>>> mod23(7).inverse()
10 (mod 23)
>>> mod23(7).inverse() * mod23(7)
1 (mod 23)


Now one very cool thing, and something that makes some basic ring theory worth understanding, is that we can compute the gcd of any number type using the exact same code for the Euclidean algorithm, provided we implement an abs function and a division algorithm. Via a chain of relatively easy-to-prove lemmas, if your number type has enough structure (in particular, if it has a division algorithm that satisfies some properties), then greatest common divisors are well-defined, and the Euclidean algorithm gives us that special linear combination. And using the same trick above in finite fields, we can use the Euclidean algorithm to compute inverses.

But in order to make things work programmatically we need to be able to deal with the literal ints 0 and 1 in the algorithm. That is, we need to be able to silently typecast integers up to whatever number type we’re working with. This makes sense because all rings have 0 and 1, but it requires a bit of scaffolding to implement. In particular, typecasting things sensibly is really difficult if you aren’t careful. And the problems are compounded in a language like Python that blatantly ignores types whenever possible. [2]

So let’s take a quick break to implement a tiny type system with implicit typecasting.

[1] The reader familiar with our series on category theory will recognize this as the product of two integers in a category whose arrows represent divisibility. So by abstract nonsense, this proves that gcd’s are unique up to multiplication by a unit in any ring.
[2] In the process of writing the code for this post, I was sorely missing the stronger type systems of Java and Haskell. Never thought I’d say that, but it’s true.

## A Tiny Generic Type System

The main driving force behind our type system will be a decorator called @typecheck. We covered decorators toward the end of our primer on dynamic programming, but in short a decorator is a Python syntax shortcut that allows some pre- or post-processing to happen to a function in a reusable way. All you need to do to apply the pre/post-processing is prefix the function definition with the name of the decorator.

Our decorator will be called typecheck, and it will decorate binary operations on our number types. In its basic form, our type checker will work as follows: if the types of the two operands are the same, then the decorator will just pass them on through to the operator. Otherwise, it will try to do some typecasting, and if that fails it will raise exceptions with reckless abandon.

def typecheck(f):
def newF(self, other):
if type(self) is not type(other):
try:
other = self.__class__(other)
except TypeError:
message = 'Not able to typecast %s of type %s to type %s in function %s'
raise TypeError(message % (other, type(other).__name__, type(self).__name__, f.__name__))
except Exception as e:
message = 'Type error on arguments %r, %r for functon %s. Reason:%s'
raise TypeError(message % (self, other, f.__name__, type(other).__name__, type(self).__name__, e))

return f(self, other)

return newF


So this is great, but there are two issues. The first is that this will only silently typecast if the thing we’re casting is on the right-hand side of the expression. In other words, the following will raise an exception complaining that you can’t add ints to Mod7 integers.

>>> x = IntegersModP(7)(1)
>>> 1 + x


What we need are the right-hand versions of all the operator overloads. They are the same as the usual operator overloads, but Python gives preference to the left-hand operator overloads. Anticipating that we will need to rewrite these silly right-hand overloads for every number type, and they’ll all be the same, we make two common base classes.

class DomainElement(object):
def __radd__(self, other): return self + other
def __rsub__(self, other): return -self + other
def __rmul__(self, other): return self * other

class FieldElement(DomainElement):
def __truediv__(self, other): return self * other.inverse()
def __rtruediv__(self, other): return self.inverse() * other
def __div__(self, other): return self.__truediv__(other)
def __rdiv__(self, other): return self.__rtruediv__(other)


And we can go ahead and make our IntegersModP a subclass of FieldElement. [3]

But now we’re wading into very deep waters. In particular, we know ahead of time that our next number type will be for Polynomials (over the integers, or fractions, or $\mathbb{Z}/p$, or whatever). And we’ll want to do silent typecasting from ints and IntegersModP to Polynomials! The astute reader will notice the discrepancy. What will happen if I try to do this?

>>> MyInteger() + MyPolynomial()


Let’s take this slowly: by our assumption both MyInteger and MyPolynomial have the __add__ and __radd__ functions defined on them, and each tries to typecast the other the appropriate type. But which is called? According to Python’s documentation if the left-hand side has an __add__ function that’s called first, and the right-hand sides’s __radd__ function is only sought if no __add__ function is found for the left operand.

Well that’s a problem, and we’ll deal with it in a half awkward and half elegant way. What we’ll do is endow our number types with an “operatorPrecedence” constant. And then inside our type checker function we’ll see if the right-hand operand is an object of higher precedence. If it is, we return the global constant NotImplemented, which Python takes to mean that no __add__ function was found, and it proceeds to look for __radd__. And so with this modification our typechecker is done. [4]

def typecheck(f):
def newF(self, other):
if (hasattr(other.__class__, 'operatorPrecedence') and
other.__class__.operatorPrecedence > self.__class__.operatorPrecedence):
return NotImplemented

if type(self) is not type(other):
try:
other = self.__class__(other)
except TypeError:
message = 'Not able to typecast %s of type %s to type %s in function %s'
raise TypeError(message % (other, type(other).__name__, type(self).__name__, f.__name__))
except Exception as e:
message = 'Type error on arguments %r, %r for functon %s. Reason:%s'
raise TypeError(message % (self, other, f.__name__, type(other).__name__, type(self).__name__, e))

return f(self, other)

return newF


We add a default operatorPrecedence of 1 to the DomainElement base class. Now this function answers our earlier question of why we want to encapsulate the prime modulus into the IntegersModP class. If this typechecker is really going to be generic, we need to be able to typecast an int by passing the single int argument to the type constructor with no additional information! Indeed, this will be the same pattern for our polynomial class and the finite field class to follow.

Now there is still one subtle problem. If we try to generate two copies of the same number type from our number-type generator (in other words, the following code snippet), we’ll get a nasty exception.

>>> mod7 = IntegersModP(7)
>>> mod7Copy = IntegersModP(7)
>>> mod7(1) + mod7Copy(2)
... fat error ...


The reason for this is that in the type-checker we’re using the Python built-in ‘is’ which checks for identity, not semantic equality. To fix this, we simply need to memoize the IntegersModP function (and all the other functions we’ll use to generate number types) so that there is only ever one copy of a number type in existence at a time.

So enough Python hacking: let’s get on with implementing finite fields!

[3] This also compels us to make some slight modifications to the constructor for IntegersModP, but they’re not significant enough to display here. Check out the Github repo if you want to see.
[4] This is truly a hack, and we’ve considered submitting a feature request to the Python devs. It is conceivably useful for the operator-overloading aficionado. I’d be interested to hear your thoughts in the comments as to whether this is a reasonable feature to add to Python.

## Polynomial Arithmetic

Recall from our finite field primer that every finite field can be constructed as a quotient of a polynomial ring with coefficients in $\mathbb{Z}/p$ by some prime ideal. We spelled out exactly what this means in fine detail in the primer, so check that out before reading on.

Indeed, to construct a finite field we need to find some irreducible monic polynomial $f$ with coefficients in $\mathbb{Z}/p$, and then the elements of our field will be remainders of arbitrary polynomials when divided by $f$. In order to determine if they’re irreducible we’ll need to compute a gcd. So let’s build a generic polynomial type with a polynomial division algorithm, and hook it into our gcd framework.

We start off in much the same way as with the IntegersModP:

# create a polynomial with coefficients in a field; coefficients are in
# increasing order of monomial degree so that, for example, [1,2,3]
# corresponds to 1 + 2x + 3x^2
@memoize
def polynomialsOver(field=fractions.Fraction):

class Polynomial(DomainElement):
operatorPrecedence = 2
factory = lambda L: Polynomial([field(x) for x in L])

def __init__(self, c):
if type(c) is Polynomial:
self.coefficients = c.coefficients
elif isinstance(c, field):
self.coefficients = [c]
elif not hasattr(c, '__iter__') and not hasattr(c, 'iter'):
self.coefficients = [field(c)]
else:
self.coefficients = c

self.coefficients = strip(self.coefficients, field(0))

def isZero(self): return self.coefficients == []

def __repr__(self):
if self.isZero():
return '0'

return ' + '.join(['%s x^%d' % (a,i) if i > 0 else '%s'%a
for i,a in enumerate(self.coefficients)])

def __abs__(self): return len(self.coefficients)
def __len__(self): return len(self.coefficients)
def __sub__(self, other): return self + (-other)
def __iter__(self): return iter(self.coefficients)
def __neg__(self): return Polynomial([-a for a in self])

def iter(self): return self.__iter__()
def degree(self): return abs(self) - 1



All of this code just lays down conventions. A polynomial is a list of coefficients (in increasing order of their monomial degree), the zero polynomial is the empty list of coefficients, and the abs() of a polynomial is one plus its degree. [5] Finally, instead of closing over a prime modulus, as with IntegersModP, we’re closing over the field of coefficients. In general you don’t have to have polynomials with coefficients in a field, but if they do come from a field then you’re guaranteed to get a sensible Euclidean algorithm. In the formal parlance, if $k$ is a field then $k[x]$ is a Euclidean domain. And for our goal of defining finite fields, we will always have coefficients from $\mathbb{Z}/p$, so there’s no problem.

Now we can define things like addition, multiplication, and equality using our typechecker to silently cast and watch for errors.

      @typecheck
def __eq__(self, other):
return self.degree() == other.degree() and all([x==y for (x,y) in zip(self, other)])

@typecheck
newCoefficients = [sum(x) for x in itertools.zip_longest(self, other, fillvalue=self.field(0))]
return Polynomial(newCoefficients)

@typecheck
def __mul__(self, other):
if self.isZero() or other.isZero():
return Zero()

newCoeffs = [self.field(0) for _ in range(len(self) + len(other) - 1)]

for i,a in enumerate(self):
for j,b in enumerate(other):
newCoeffs[i+j] += a*b

return Polynomial(newCoeffs)


Notice that, if the underlying field of coefficients correctly implements the operator overloads, none of this depends on the coefficients. Reusability, baby!

And we can finish off with the division algorithm for polynomials.

      @typecheck
def __divmod__(self, divisor):
quotient, remainder = Zero(), self
divisorDeg = divisor.degree()

while remainder.degree() >= divisorDeg:
monomialExponent = remainder.degree() - divisorDeg
monomialZeros = [self.field(0) for _ in range(monomialExponent)]
monomialDivisor = Polynomial(monomialZeros + [remainder.leadingCoefficient() / divisorLC])

quotient += monomialDivisor
remainder -= monomialDivisor * divisor

return quotient, remainder


Indeed, we’re doing nothing here but formalizing the grade-school algorithm for doing polynomial long division [6]. And we can finish off the function for generating this class by assigning the field member variable along with a class name. And we give it a higher operator precedence than the underlying field of coefficients so that an isolated coefficient is cast up to a constant polynomial.

@memoize
def polynomialsOver(field=fractions.Fraction):

class Polynomial(DomainElement):
operatorPrecedence = 2

[... methods defined above ...]

def Zero():
return Polynomial([])

Polynomial.field = field
Polynomial.__name__ = '(%s)[x]' % field.__name__
return Polynomial


We provide a modest test suite in the Github repository for this post, but here’s a sample test:

>>> Mod5 = IntegersModP(5)
>>> Mod11 = IntegersModP(11)
>>> polysOverQ = polynomialsOver(Fraction).factory
>>> polysMod5 = polynomialsOver(Mod5).factory
>>> polysMod11 = polynomialsOver(Mod11).factory
>>> polysOverQ([1,7,49]) / polysOverQ([7])
1/7 + 1 x^1 + 7 x^2
>>> polysMod5([1,7,49]) / polysMod5([7])
3 + 1 x^1 + 2 x^2
>>> polysMod11([1,7,49]) / polysMod11([7])
8 + 1 x^1 + 7 x^2


And indeed, the extended Euclidean algorithm works without modification, so we know our typecasting is doing what’s expected:

>>> p = polynomialsOver(Mod2).factory
>>> f = p([1,0,0,0,1,1,1,0,1,1,1]) # x^10 + x^9 + x^8 + x^6 + x^5 + x^4 + 1
>>> g = p([1,0,1,1,0,1,1,0,0,1])   # x^9 + x^6 + x^5 + x^3 + x^1 + 1
>>> theGcd = p([1,1,0,1]) # x^3 + x + 1
>>> x = p([0,0,0,0,1]) # x^4
>>> y = p([1,1,1,1,1,1]) # x^5 + x^4 + x^3 + x^2 + x + 1
>>> (x,y,theGcd) == extendedEuclideanAlgorithm(f, g)
True


[5] The mathematical name for the abs() function that we’re using is a valuation.
[6] One day we will talk a lot more about polynomial long division on this blog. You can do a lot of cool algebraic geometry with it, and the ideas there lead you to awesome applications like robot motion planning and automated geometry theorem proving.

## Generating Irreducible Polynomials

Now that we’ve gotten Polynomials out of the way, we need to be able to generate irreducible polynomials over $\mathbb{Z}/p$ of any degree we want. It might be surprising that irreducible polynomials of any degree exist [7], but in fact we know a lot more.

Theorem: The product of all irreducible monic polynomials of degree dividing $m$ is equal to $x^{p^m} - x$.

This is an important theorem, but it takes a little bit more field theory than we have under our belts right now. You could summarize the proof by saying there is a one-to-one correspondence between elements of a field and monic irreducible polynomials, and then you say some things about splitting fields. You can see a more detailed proof outline here, but it assumes you’re familiar with the notion of a minimal polynomial. We will probably cover this in a future primer.

But just using the theorem we can get a really nice algorithm for determining if a polynomial $f(x)$ of degree $m$ is irreducible: we just look at its gcd with all the $x^{p^k} - x$ for $k$ smaller than $m$. If all the gcds are constants, then we know it’s irreducible, and if even one is a non-constant polynomial then it has to be irreducible. Why’s that? Because if you have some nontrivial gcd $d(x) = \gcd(f(x), x^{p^k} - x)$ for $k < m$, then it’s a factor of $f(x)$ by definition. And since we know all irreducible monic polynomials are factors of that this collection of polynomials, if the gcd is always 1 then there are no other possible factors to be divisors. (If there is any divisor then there will be a monic irreducible one!) So the candidate polynomial must be irreducible. In fact, with a moment of thought it’s clear that we can stop at $k= m/2$, as any factor of large degree will necessarily require corresponding factors of small degree. So the algorithm to check for irreducibility is just this simple loop:

def isIrreducible(polynomial, p):
ZmodP = IntegersModP(p)
poly = polynomialsOver(ZmodP).factory
x = poly([0,1])
powerTerm = x
isUnit = lambda p: p.degree() == 0

for _ in range(int(polynomial.degree() / 2)):
powerTerm = powerTerm.powmod(p, polynomial)
gcdOverZmodp = gcd(polynomial, powerTerm - x)
if not isUnit(gcdOverZmodp):
return False

return True


We’re just computing the powers iteratively as $x^p, (x^p)^p = x^{p^2}, \dots, x^{p^j}$ and in each step of the loop subtracting $x$ and computing the relevant gcd. The powmod function is just there so that we can reduce the power mod our irreducible polynomial after each multiplication, keeping the degree of the polynomial small and efficient to work with.

Now generating an irreducible polynomial is a little bit harder than testing for one. With a lot of hard work, however, field theorists discovered that irreducible polynomials are quite common. In fact, if you just generate the coefficients of your degree $n$ monic polynomial at random, the chance that you’ll get something irreducible is at least $1/n$. So this suggests an obvious randomized algorithm: keep guessing until you find one.

def generateIrreduciblePolynomial(modulus, degree):
Zp = IntegersModP(modulus)
Polynomial = polynomialsOver(Zp)

while True:
coefficients = [Zp(random.randint(0, modulus-1)) for _ in range(degree)]
randomMonicPolynomial = Polynomial(coefficients + [Zp(1)])

if isIrreducible(randomMonicPolynomial, modulus):
return randomMonicPolynomial


Since the probability of getting an irreducible polynomial is close to $1/n$, we expect to require $n$ trials before we find one. Moreover we could give a pretty tight bound on how likely it is to deviate from the expected number of trials. So now we can generate some irreducible polynomials!

>>> F23 = FiniteField(2,3)
>>> generateIrreduciblePolynomial(23, 3)
21 + 12 x^1 + 11 x^2 + 1 x^3



And so now we are well-equipped to generate any finite field we want! It’s just a matter of generating the polynomial and taking a modulus after every operation.

@memoize
def FiniteField(p, m, polynomialModulus=None):
Zp = IntegersModP(p)
if m == 1:
return Zp

Polynomial = polynomialsOver(Zp)
if polynomialModulus is None:
polynomialModulus = generateIrreduciblePolynomial(modulus=p, degree=m)

class Fq(FieldElement):
fieldSize = int(p ** m)
primeSubfield = Zp
idealGenerator = polynomialModulus
operatorPrecedence = 3

def __init__(self, poly):
if type(poly) is Fq:
self.poly = poly.poly
elif type(poly) is int or type(poly) is Zp:
self.poly = Polynomial([Zp(poly)])
elif isinstance(poly, Polynomial):
self.poly = poly % polynomialModulus
else:
self.poly = Polynomial([Zp(x) for x in poly]) % polynomialModulus

self.field = Fq

@typecheck
def __add__(self, other): return Fq(self.poly + other.poly)
@typecheck
def __sub__(self, other): return Fq(self.poly - other.poly)
@typecheck
def __mul__(self, other): return Fq(self.poly * other.poly)
@typecheck
def __eq__(self, other): return isinstance(other, Fq) and self.poly == other.poly

def __pow__(self, n): return Fq(pow(self.poly, n))
def __neg__(self): return Fq(-self.poly)
def __abs__(self): return abs(self.poly)
def __repr__(self): return repr(self.poly) + ' \u2208 ' + self.__class__.__name__

@typecheck
def __divmod__(self, divisor):
q,r = divmod(self.poly, divisor.poly)
return (Fq(q), Fq(r))

def inverse(self):
if self == Fq(0):
raise ZeroDivisionError

x,y,d = extendedEuclideanAlgorithm(self.poly, self.idealGenerator)
return Fq(x) * Fq(d.coefficients[0].inverse())

Fq.__name__ = 'F_{%d^%d}' % (p,m)
return Fq


And some examples of using it:

>>> F23 = FiniteField(2,3)
>>> x = F23([1,1])
>>> x
1 + 1 x^1 ∈ F_{2^3}
>>> x*x
1 + 0 x^1 + 1 x^2 ∈ F_{2^3}
>>> x**10
0 + 0 x^1 + 1 x^2 ∈ F_{2^3}
>>> 1 / x
0 + 1 x^1 + 1 x^2 ∈ F_{2^3}
>>> x * (1 / x)
1 ∈ F_{2^3}

>>> k = FiniteField(23, 4)
>>> k.fieldSize
279841
>>> k.idealGenerator
6 + 8 x^1 + 10 x^2 + 10 x^3 + 1 x^4
>>> y
9 + 21 x^1 + 14 x^2 + 12 x^3 ∈ F_{23^4}
>>> y*y
13 + 19 x^1 + 7 x^2 + 14 x^3 ∈ F_{23^4}
>>> y**5 - y
15 + 22 x^1 + 15 x^2 + 5 x^3 ∈ F_{23^4}


And that’s it! Now we can do arithmetic over any finite field we want.

[7] Especially considering that other wacky things happen like this: $x^4 +1$ is reducible over every finite field!

## Some Words on Efficiency

There are a few things that go without stating about this program, but I’ll state them anyway.

The first is that we pay a big efficiency price for being so generic. All the typecasting we’re doing isn’t cheap, and in general cryptography needs to be efficient. For example, if I try to create a finite field of order $104729^{20}$, it takes about ten to fifteen seconds to complete. This might not seem so bad for a one-time initialization, but it’s clear that our representation is somewhat bloated. We would display a graph of the expected time to perform various operations in various finite fields, but this post is already way too long.

In general, the larger and more complicated the polynomial you use to define your finite field, the longer operations will take (since dividing by a complicated polynomial is more expensive than dividing by a simple polynomial). For this reason and a few other reasons, a lot of research has gone into efficiently finding irreducible polynomials with, say, only three nonzero terms. Moreover, if we know in advance that we’ll only work over fields of characteristic two we can make a whole lot of additional optimizations. Essentially, all of the arithmetic reduces to really fast bitwise operations, and things like exponentiation can easily be implemented in hardware. But it also seems that the expense coming with large field characteristics corresponds to higher security, so some researchers have tried to meet in the middle an get efficient representations of other field characteristics.

In any case, the real purpose of our implementation in this post is not for efficiency. We care first and foremost about understanding the mathematics, and to have a concrete object to play with and use in the future for other projects. And we have accomplished just that.

Until next time!

# (Finite) Fields — A Primer

So far on this blog we’ve given some introductory notes on a few kinds of algebraic structures in mathematics (most notably groups and rings, but also monoids). Fields are the next natural step in the progression.

If the reader is comfortable with rings, then a field is extremely simple to describe: they’re just commutative rings with 0 and 1, where every nonzero element has a multiplicative inverse. We’ll give a list of all of the properties that go into this “simple” definition in a moment, but an even more simple way to describe a field is as a place where “arithmetic makes sense.” That is, you get operations for $+,-, \cdot , /$ which satisfy the expected properties of addition, subtraction, multiplication, and division. So whatever the objects in your field are (and sometimes they are quite weird objects), they behave like usual numbers in a very concrete sense.

So here’s the official definition of a field. We call a set $F$ a field if it is endowed with two binary operations addition ($+$) and multiplication ($\cdot$, or just symbol juxtaposition) that have the following properties:

• There is an element we call 0 which is the identity for addition.
• Addition is commutative and associative.
• Every element $a \in F$ has a corresponding additive inverse $b$ (which may equal $a$) for which $a + b = 0$.

These three properties are just the axioms of a (commutative) group, so we continue:

• There is an element we call 1 (distinct from 0) which is the identity for multiplication.
• Multiplication is commutative and associative.
• Every nonzero element $a \in F$ has a corresponding multiplicative inverse $b$ (which may equal $a$) for which $ab = 1$.
• Addition and multiplication distribute across each other as we expect.

If we exclude the existence of multiplicative inverses, these properties make $F$ a commutative ring, and so we have the following chain of inclusions that describes it all

$\displaystyle \textup{Fields} \subset \textup{Commutative Rings} \subset \textup{Rings} \subset \textup{Commutative Groups} \subset \textup{Groups}$

The standard examples of fields are the real numbers $\mathbb{R}$, the rationals $\mathbb{Q}$, and the complex numbers $\mathbb{C}$. But of course there are many many more. The first natural question to ask about fields is: what can they look like?

For example, can there be any finite fields? A field $F$ which as a set has only finitely many elements?

As we saw in our studies of groups and rings, the answer is yes! The simplest example is the set of integers modulo some prime $p$. We call them $\mathbb{Z} / p \mathbb{Z},$ or sometimes just $\mathbb{Z}/p$ for short, and let’s rederive what we know about them now.

As a set, $\mathbb{Z}/p$ consists of the integers $\left \{ 0, 1, \dots, p-1 \right \}$. The addition and multiplication operations are easy to define, they’re just usual addition and multiplication followed by a modulus. That is, we add by $a + b \mod p$ and multiply with $ab \mod p$. This thing is clearly a commutative ring (because the integers form a commutative ring), so to show this is a field we need to show that everything has a multiplicative inverse.

There is a nice fact that allows us to do this: an element $a$ has an inverse if and only if the only way for it to divide zero is the trivial way $0a = 0$. Here’s a proof. For one direction, suppose $a$ divides zero nontrivially, that is there is some $c \neq 0$ with $ac = 0$. Then if $a$ had an inverse $b$, then $0 = b(ac) = (ba)c = c$, but that’s very embarrassing for $c$ because it claimed to be nonzero. Now suppose $a$ only divides zero in the trivial way. Then look at all possible ways to multiply $a$ by other nonzero elements of $F$. No two can give you the same result because if $ax = ay$ then (without using multiplicative inverses) $a(x-y) = 0$, but we know that $a$ can only divide zero in the trivial way so $x=y$. In other words, the map “multiplication by $a$” is injective. Because the set of nonzero elements of $F$ is finite you have to hit everything (the map is in fact a bijection), and some $x$ will give you $ax = 1$.

Now let’s use this fact on $\mathbb{Z}/p$ in the obvious way. Since $p$ is a prime, there are no two smaller numbers $a, b < p$ so that $ab = p$. But in $\mathbb{Z}/p$ the number $p$ is equivalent to zero (mod $p$)! So $\mathbb{Z}/p$ has no nontrivial zero divisors, and so every element has an inverse, and so it’s a finite field with $p$ elements.

The next question is obvious: can we get finite fields of other sizes? The answer turns out to be yes, but you can’t get finite fields of any size. Let’s see why.

## Characteristics and Vector Spaces

Say you have a finite field $k$ (lower-case k is the standard letter for a field, so let’s forget about $F$). Beacuse the field is finite, if you take 1 and keep adding it to itself you’ll eventually run out of field elements. That is, $n = 1 + 1 + \dots + 1 = 0$ at some point. How do I know it’s zero and doesn’t keep cycling never hitting zero? Well if at two points $n = m \neq 0$, then $n-m = 0$ is a time where you hit zero, contradicting the claim.

Now we define $\textup{char}(k)$, the characteristic of $k$, to be the smallest $n$ (sums of 1 with itself) for which $n = 0$. If there is no such $n$ (this can happen if $k$ is infinite, but doesn’t always happen for infinite fields), then we say the characteristic is zero. It would probably make more sense to say the characteristic is infinite, but that’s just the way it is. Of course, for finite fields the characteristic is always positive. So what can we say about this number? We have seen lots of example where it’s prime, but is it always prime? It turns out the answer is yes!

For if $ab = n = \textup{char}(k)$ is composite, then by the minimality of $n$ we get $a,b \neq 0$, but $ab = n = 0$. This can’t happen by our above observation, because being a zero divisor means you have no inverse! Contradiction, sucker.

But it might happen that there are elements of $k$ that can’t be written as $1 + 1 + \dots + 1$ for any number of terms. We’ll construct examples in a minute (in fact, we’ll classify all finite fields), but we already have a lot of information about what those fields might look like. Indeed, since every field has 1 in it, we just showed that every finite field contains a smaller field (a subfield) of all the ways to add 1 to itself. Since the characteristic is prime, the subfield is a copy of $\mathbb{Z}/p$ for $p = \textup{char}(k)$. We call this special subfield the prime subfield of $k$.

The relationship between the possible other elements of $k$ and the prime subfield is very neat. Because think about it: if $k$ is your field and $F$ is your prime subfield, then the elements of $k$ can interact with $F$ just like any other field elements. But if we separate $k$ from $F$ (make a separate copy of $F$), and just think of $k$ as having addition, then the relationship with $F$ is that of a vector space! In fact, whenever you have two fields $k \subset k'$, the latter has the structure of a vector space over the former.

Back to finite fields, $k$ is a vector space over its prime subfield, and now we can impose all the power and might of linear algebra against it. What’s it’s dimension? Finite because $k$ is a finite set! Call the dimension $m$, then we get a basis $v_1, \dots, v_m$. Then the crucial part: every element of $k$ has a unique representation in terms of the basis. So they are expanded in the form

$\displaystyle f_1v_1 + \dots + f_mv_m$

where the $f_i$ come from $F$. But now, since these are all just field operations, every possible choice for the $f_i$ has to give you a different field element. And how many choices are there for the $f_i$? Each one has exactly $|F| = \textup{char}(k) = p$. And so by counting we get that $k$ has $p^m$ many elements.

This is getting exciting quickly, but we have to pace ourselves! This is a constraint on the possible size of a finite field, but can we realize it for all choices of $p, m$? The answer is again yes, and in the next section we’ll see how.  But reader be warned: the formal way to do it requires a little bit of familiarity with ideals in rings to understand the construction. I’ll try to avoid too much technical stuff, but if you don’t know what an ideal is, you should expect to get lost (it’s okay, that’s the nature of learning new math!).

## Constructing All Finite Fields

Let’s describe a construction. Take a finite field $k$ of characteristic $p$, and say you want to make a field of size $p^m$. What we need to do is construct a field extension, that is, find a bigger field containing $k$ so that the vector space dimension of our new field over $k$ is exactly $m$.

What you can do is first form the ring of polynomials with coefficients in $k$. This ring is usually denoted $k[x]$, and it’s easy to check it’s a ring (polynomial addition and multiplication are defined in the usual way). Now if I were speaking to a mathematician I would say, “From here you take an irreducible monic polynomial $p(x)$ of degree $m$, and quotient your ring by the principal ideal generated by $p$. The result is the field we want!”

In less compact terms, the idea is exactly the same as modular arithmetic on integers. Instead of doing arithmetic with integers modulo some prime (an irreducible integer), we’re doing arithmetic with polynomials modulo some irreducible polynomial $p(x)$. Now you see the reason I used $p$ for a polynomial, to highlight the parallel thought process. What I mean by “modulo a polynomial” is that you divide some element $f$ in your ring by $p$ as much as you can, until the degree of the remainder is smaller than the degree of $p(x)$, and that’s the element of your quotient. The Euclidean algorithm guarantees that we can do this no matter what $k$ is (in the formal parlance, $k[x]$ is called a Euclidean domain for this very reason). In still other words, the “quotient structure” tells us that two polynomials $f, g \in k[x]$ are considered to be the same in $k[x] / p$ if and only if $f - g$ is divisible by $p$. This is actually the same definition for $\mathbb{Z}/p$, with polynomials replacing numbers, and if you haven’t already you can start to imagine why people decided to study rings in general.

Let’s do a specific example to see what’s going on. Say we’re working with $k = \mathbb{Z}/3$ and we want to compute a field of size $27 = 3^3$. First we need to find a monic irreducible polynomial of degree $3$. For now, I just happen to know one: $p(x) = x^3 - x + 1$. In fact, we can check it’s irreducible, because to be reducible it would have to have a linear factor and hence a root in $\mathbb{Z}/3$. But it’s easy to see that if you compute $p(0), p(1), p(2)$ and take (mod 3) you never get zero.

So I’m calling this new ring

$\displaystyle \frac{\mathbb{Z}/3[x]}{(x^3 - x + 1)}$

It happens to be a field, and we can argue it with a whole lot of ring theory. First, we know an irreducible element of this ring is also prime (because the ring is a unique factorization domain), and prime elements generate maximal ideals (because it’s a principal ideal domain), and if you quotient by a maximal ideal you get a field (true of all rings).

But if we want to avoid that kind of argument and just focus on this ring, we can explicitly construct inverses. Say you have a polynomial $f(x)$, and for illustration purposes we’ll choose $f(x) = x^4 + x^2 - 1$. Now in the quotient ring we could do polynomial long division to find remainders, but another trick is just to notice that the quotient is equivalent to the condition that $x^3 = x - 1$. So we can reduce $f(x)$ by applying this rule to $x^4 = x^3 x$ to get

$\displaystyle f(x) = x^2 + x(x-1) - 1 = 2x^2 - x - 1$

Now what’s the inverse of $f(x)$? Well we need a polynomial $g(x) = ax^2 + bx + c$ whose product with $f$ gives us something which is equivalent to 1, after you reduce by $x^3 - x + 1$. A few minutes of algebra later and you’ll discover that this is equivalent to the following polynomial being identically 1

$\displaystyle (a-b+2c)x^2 + (-3a+b-c)x + (a - 2b - 2c) = 1$

In other words, we get a system of linear equations which we need to solve:

\displaystyle \begin{aligned} a & - & b & + & 2c & = 0 \\ -3a & + & b & - & c &= 0 \\ a & - & 2b & - & 2c &= 1 \end{aligned}

And from here you can solve with your favorite linear algebra techniques. This is a good exercise for working in fields, because you get to abuse the prime subfield being characteristic 3 to say terrifying things like $-1 = 2$ and $6b = 0$. The end result is that the inverse polynomial is $2x^2 + x + 1$, and if you were really determined you could write a program to compute these linear systems for any input polynomial and ensure they’re all solvable. We prefer the ring theoretic proof.

In any case, it’s clear that taking a polynomial ring like this and quotienting by a monic irreducible polynomial gives you a field. We just control the size of that field by choosing the degree of the irreducible polynomial to our satisfaction. And that’s how we get all finite fields!

## One Last Word on Irreducible Polynomials

One thing we’ve avoided is the question of why irreducible monic polynomials exist of all possible degrees $m$ over any $\mathbb{Z}/p$ (and as a consequence we can actually construct finite fields of all possible sizes).

The answer requires a bit of group theory to prove this, but it turns out that the polynomial $x^{p^m} - x$ has all degree $m$ monic irreducible polynomials as factors. But perhaps a better question (for computer scientists) is how do we work over a finite field in practice? One way is to work with polynomial arithmetic as we described above, but this has some downsides: it requires us to compute these irreducible monic polynomials (which doesn’t sound so hard, maybe), to do polynomial long division every time we add, subtract, or multiply, and to compute inverses by solving a linear system.

But we can do better for some special finite fields, say where the characteristic is 2 (smells like binary) or we’re only looking at $F_{p^2}$. The benefit there is that we aren’t forced to use polynomials. We can come up with some other kind of structure (say, matrices of a special form) which happens to have the same field structure and makes computing operations relatively painless. We’ll see how this is done in the future, and see it applied to cryptography when we continue with our series on elliptic curve cryptography.

Until then!