Last week, we looked at the finite fields of prime order, Fp. A field, remember, is a set in which you can do addition, subtraction, multiplication and division. For a given prime p, the field Fp consists of the set {0, 1, ..., p-1}, with arithmetic operations done modulo p. Why does p have to be a prime? We can do addition, subtraction, and multiplication modulo n for any n, prime or not. However, for division, we require p to be prime. (Why?)

You might think, therefore, that the fields Fp of prime order are the only finite fields. However, you would be wrong. We shall see that there are in fact fields Fq of order q (ie, with q elements) for every prime power q = p^n. However, before we can do that we need to understand about algebraic extensions of a field.

It can happen that a field is contained within another field. For example, we have Q

- At step 0, start with Q
- At step 1, add a
- At step n, add all elements that can be obtained by combining two elements from step n-1 using one of the arithmetic operators.

The field Q(a), obtained by adjoining a single element, is called a simple extension of Q. (We could then go on to adjoin another element b, to get Q(a,b), and so on.) Among simple extensions, there are two ways it can go.

Suppose that a is a zero of some polynomial with coefficients in Q - for example x^2-2. In that case, we will construct the whole of Q(a) after only a finite number of steps. For example, in the case where a = sqrt 2, every element of Q(sqrt 2) can be expressed as c + d sqrt 2, for some c, d in Q. For example:

(sqrt 2)^2 = 2

1/(sqrt 2) = (sqrt 2) / 2

1/(1 + sqrt 2) = (1 - sqrt 2) / (1 - 2) = -1 + sqrt 2

This is called an algebraic extension.

Alternatively, if a is not the zero of a polynomial over Q (for example: e, pi), then the construction of Q(a) in steps will never finish. This is called a transcendental extension.

Algebraic extensions of Q are fascinating things. Initially, the main motivation for studying them came from number theory. Some other time, I'd like to take a closer look at them. However, for the moment, I just want to show how to construct them in Haskell, as what we're really aiming for is algebraic extensions of the finite fields Fp.

Let's suppose that we're trying to construct Q(sqrt2). The polynomial we're interested in is x^2-2. Then the basic idea is:

- Form the polynomial ring Q[x], consisting of polynomials in x with coefficients in Q
- Represent elements of Q(a) as polynomials in Q[x]
- Do addition, subtraction, multiplication in Q(a) using the underlying operations in Q[x]
- After any arithmetic operation, replace the result by its remainder on division by x^2-2.

To construct Q(sqrt3), Q(i), etc, just replace x^2-2 by the appropriate polynomial.

Okay, rather belatedly, time for some code. First, we need to define a type for (univariate) polynomials:

newtype UPoly a = UP [a] deriving (Eq,Ord)

UP [c0,c1,...cn] is to be interpreted as the polynomial c0 + c1 x + ... + cn x^n.

Exercise: Define a Num instance for UPoly a.

If we then define

x = UP [0,1] :: UPoly Integer

together with a suitable Show instance, then we can do things like the following:

> (1+x)^3

1+3x+3x^2+x^3

Exercise: Write quotRemUP, to perform division with remainder in UPoly k, on the assumption that k is a field (that is, a Fractional instance).

So we now have a type, UPoly Q, representing Q[x]. Next, we want to wrap this in another type to represent extension fields Q(a). Rather than have to do this over again each time for Q(sqrt2), Q(sqrt3), Q(i), and so on, we use a little bit of phantom type trickery again. Last time, we used phantom types to represent integers; this time, we're going to use phantom types to represent the polynomials that define the fields (x^2-2, x^2-3, x^2+1, etc).

class PolynomialAsType k poly where

pvalue :: (k,poly) -> UPoly k

data ExtensionField k poly = Ext (UPoly k) deriving (Eq,Ord)

Here, k represents the field we are extending - Q, to begin with - and if a is the element that we want to adjoin, then poly represents the polynomial over Q of which a is a zero.

From here, it's a short step to define some extension fields:

data Sqrt a = Sqrt a

-- n should be square-free

instance IntegerAsType n => PolynomialAsType Q (Sqrt n) where

pvalue _ = convert $ x^2 - fromInteger (value (undefined :: n))

type QSqrt2 = ExtensionField Q (Sqrt T2)

sqrt2 = embed x :: QSqrt2

type QSqrt3 = ExtensionField Q (Sqrt T3)

sqrt3 = embed x :: QSqrt3

And now we can do arithmetic in Q(sqrt2), Q(sqrt3), etc. For example:

> :set +t

> (1+sqrt2)^2

3+2a :: QSqrt2

> (1+sqrt3)^2

4+2a :: QSqrt3

As you see, the show function of ExtensionField k poly shows the adjoined element as "a", regardless of which field we're working in. With a little more type hackery, we could have made that a parameter to the type too, so that it would show "sqrt2", "sqrt3", "i", depending which field we were in. However, that would have obscured the code even more. In practice, when we use this code we're only going to be working over one field at a time, so it's fine as it is.

Another limitation of this code is that it's going to be a bit unwieldy to construct Q(a,b) as the type ExtensionField (ExtensionField k poly1) poly2. Luckily, this doesn't matter, as any algebraic extension Q(a,b) is equal to Q(c) for some c.

Exercise: Show that Q(sqrt2, sqrt3) = Q(sqrt2 + sqrt3), and find the polynomial of which sqrt2 + sqrt3 is a zero.

I apologise that I've only really explained the bare bones of how this works. Hopefully you can fill in the gaps. (They're all in the HaskellForMaths source - see link at top of page.) This post has already taken me far too long to write, but there is just one other thing I ought to mention.

Field extensions can have automorphisms (symmetries). Recall that a symmetry is a change that leaves something looking the same. In the case of Q(sqrt2), there is a non-trivial symmetry that sends c + d sqrt 2 to c - d sqrt2 (conjugation). Field automorphisms have a very important place in the history of mathematics: Galois invented group theory in order to study these symmetries, during his investigations into the unsolvability (in general) of the quintic.

Hi,

ReplyDeleteLuckily, this doesn't matter, as any algebraic extension Q(a,b) is equal to Q(c) for some c.

How does one compute such a c?

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Shawn,

ReplyDeleteUnfortunately I can't find my reference for this claim just now. If I could, the proof would probably be constructive. In the case of Q(sqrt2, sqrt3), Q(sqrt2 + sqrt3) does the trick. If you work through that example, it might all become clear.

(Clearly sqrt2 + sqrt3 is in Q(sqrt2, sqrt3). We have to show the other way round, that sqrt2, sqrt3 are in Q(sqrt2 + sqrt3). Well, consider (sqrt2 + sqrt3)^2, ^3, etc.)

The claim is generally known as "Primitive Element Theorem". It has a half-way constructive proof.

ReplyDeletehttp://en.wikipedia.org/wiki/Primitive_element_theorem