"The time has come," the Walrus said,
"To talk of many things:
Of shoes--and ships--and sealing-wax--
Of cabbages--and rings
Well, that's not exactly what he said, but close enough.
In our
last installment of the series on algebraic number theory, we reviewed a couple of very important elementary examples of rings, namely the integers ℤ and the finite "quotient rings" ℤ/nℤ for any integer n>1. Earlier (
here) we provided the basic abstract algebraic definition of a ring. In this and subsequent installments we're going to tackle rings in earnest, as they provide the best way to formalize the bulk of concepts needed to discuss algebraic number theory seriously.
Recall that our original objective was to find solutions to polynomial equations f(x)=0, and usually more specifically where the coefficients of the polynomial are ordinary integers in ℤ. Even more specifically, we were interested in
Diophantine equations, where we wanted
solutions that are integers, or "nearly" so.
Abstractly, we know by the fundamental theorem of algebra that if n is the degree of f(x), then exactly (counting multiplicity) n solutions exist in the complex numbers ℂ. If the coefficients of f(x) were rational numbers (in ℚ), then these solutions are by definition algebraic numbers. We know, further, that in some sense
Galois theory provides a good description (so far as one can be given) of the nature of solutions of one polynomial equation in one variable with rational coefficients.
What we might hope for is finding solutions of such equations in integer or rational numbers. But that's very hard, so pragmatically it is best to ease off a bit and look for solutions in a somewhat larger class that's easier to handle theoretically. The key to the whole enterprise is finding the right abstraction to work with. We need to seek solutions which are conceptually somewhere between integers and fully general algebraic numbers. If we can deal with some intermediate construct, hopefully we can (with care) manage to bridge the gap and cross the chasm.
So let's call the concept we are looking for
algebraic integers, and try to figure out how this concept should be defined. We have an advantage over mathematicians of the 19
th century who first thought about these issues. Namely, and as a direct result of their work, we now have a collection of abstract algebraic concepts that have proven very useful for organizing the theory. High in importance among these concepts is that of a ring. Since ℤ is a ring, if algebraic integers are to form a construct that generalizes ℤ, this construct had better be a ring if we dare call it a generalization.
The next point to consider is that what is to be treated as an algebraic integer depends very much on being relative to a specific finite extension F⊇ℚ. More explicitly, if F is a field which is a finite extension of ℚ (that is, has finite degree over ℚ), F is called a
number field or
algebraic number field, because all of its members are algebraic numbers as defined previously (roots of some polynomial f(x)∈ℚ[x]).
To give a name to that which we are looking to define, let O
F (or O
F/ℚ when we want to be explicit about the base field) be the set of all elements of F that are algebraic integers. In short, F is a field of algebraic numbers over ℚ, and O
F is to be a ring that plays a role in F analogous to the role of ℤ in ℚ. So we need to have ℤ⊆ O
F.
For a clue as to how to define O
F, suppose F=ℚ. F could then be defined as the set of solutions of f(x)=0, where f(x) is a first degree polynomial with coefficients in ℤ. That is, f(x) has the form ax+b, for a,b∈ℤ and a≠0. In other words, F is just all fractions b/a with a,b∈ℤ and a≠0, namely ℚ.
In this trivial case, how are the integers defined? All we need to do is require that f(x) be a
monic first degree polynomial with coefficients in ℤ, that is f(x) has the form x+a for a∈ℤ. This suggests defining O
F as the set of elements of F that have a minimal polynomial f(x) which is monic and has coefficients in ℤ. Somewhat amazingly, this turns out to work nicely.
With this definition, ℤ⊆O
F obviously. What isn't so obvious and needs to be shown is that O
F is a ring. In particular, if a,b∈O
F, then both a+b and ab are also in O
F. That's a little work, but not hard. It can even be shown that F is the "field of fractions" of O
F, namely the set of all quotients b/a with a,b∈O
F and a≠0. Beyond that, there are a whole slew of other results that flow from this definition and justify it many times over.
So rings of integers as defined here are the natural generalization of the rational integers ℤ, which is a subring of any ring of algebraic integers. It is fair to say that the main concern of algebraic number theory is determining properties of such rings O
F for algebraic number fields F. However, there are various properties ℤ has which general rings of integers do
not have. For example, all ideals of ℤ are "principal" ideals, and all elements of ℤ factor uniquely as a product of primes. Most rings of algebraic integers have neither of these properties, but criteria can be given for when they are present.
For future reference, note that we can also talk about rings of integers of arbitrary algebraic extensions E⊇F, where the base field isn't necessarily ℚ. The definition of the integers of E/F, denoted by O
E/F, is simply all elements of E having a minimal polynomial over F which is monic and has coefficients in O
F. This is useful for the general theory, but harder to work with when developing the basic theory.
To conclude this installment, let's look at the simplest sort of extension of ℚ, a quadratic extension F=ℚ(√d), where d is a square-free integer, which may be either positive or negative. Now √d is an algebraic integer because it satisfies x
2-d=0. So is, for any b∈ℤ, b√d, since it satisfies x
2-db
2=0.
However, the situation for a+b√d, a,b∈ℤ isn't quite as obvious, unless we use the general fact (that hasn't been proven here) that the algebraic integers of F⊇ℚ form a ring, so that sums of integers are integers. It would be clearer just to produce the equation a+b√d satisfies in order to see directly that it is an integer. Fortunately this is quite easy.
Let α=a+b√d. From Galois theory, we recall that the "conjugate" of α is α*=a-b√d, and we have (x-α)(x-α*)=0. Multiplying things out we have f(α)=0, where f(x)=x
2-2ax+a
2-db
2. Clearly f(x) is monic with integer coefficients, so a+b√d is an algebraic integer in ℚ(√d).
You might be tempted to conclude that O
F is {a+b√d | a,b∈ℤ}, no matter what d is. But that's not true either. For example, if d=5, you can easily check that α=(1+√5)/2 satisfies x
2-x-1=0, so α is an algebraic integer of ℚ(√5).
What actually is true is that O
F = {a+b√d | a,b∈ℤ} if d is square-free and d ≡ 2 or 3 (mod 4). However, if d ≡ 1 (mod 4), then O
F is {(a+b√d)/2 | a,b∈ℤ}. The proof isn't hard, and we'll come back to it later.
In the next installment we'll look further into some of the ring theory relevant to rings of integers.
Tags:
algebraic number theory,
algebraic integers
