First, I forgot to add to the previous algebra post the following conclusion: if K is a number field, then the set of all algebraic integers in K is a ring. That follows directly from the fact that B is a ring – if a and b are in O(K), then a+b, a–b, and ab will be algebraic integers as well, and will be in K, so they’ll be in O(K).
The ultimate purpose of this subseries of my algebra posts is to show that for every K, O(K) has an integral basis. In other words, if K has dimension n over Q, then there exist n elements in O(K) that are linearly independent, and span O(K) over Z. Once I show that, showing that O(K) is always a Dedekind domain is almost trivial (the key fact is that every ideal has finite index – that is, O(K)/I has finitely many elements).
But it’s not the most boring part of algebraic number theory for nothing, which means there are still some more things to talk about.
First, every algebraic number a has something called a minimal polynomial. There are two ways to define it. First, F[x] is a PID for every field F – in fact, it’s a Euclidean domain, with the Euclidean function being the degree of the polynomial. Since a is algebraic, it’s the root of some nonzero polynomial in Q[x], so the ideal of all polynomials that are 0 at a is generated by some nonzero polynomial, call it m(a)(x). We can multiply m(a) by any constant we like, so we normalize it to be monic. It’s then possible to show that a is in B iff m(a) has integer coefficients.
Alternatively, if f(a) = 0, and f = gh, then g(a)h(a) = 0, so a is a root of g or h. We can continue to divide like this until we get an irreducible polynomial. That irreducible polynomial will be m(a), because the ideal it generates is maximal in Q[x], so either a is the root of every polynomial, which is impossible, or a is a root only of polynomials in the ideal. Also, note that this shows that Q[a] = Q[x]/(m(a)) is a field.
Another key fact is that every number field K can be written as Q(a) for some algebraic number a. We can improve this by forcing a to be an algebraic integer. To see why that works, let k be an integer large enough to be divisible by all denominators of coefficients of m(a), and fix b = ak. A natural function that has b as a root is then m(a)(x/k). If n is the degree of m(a), then we need to multiply m(a)(x/k) by k^n to make it a monic polynomial. But then the coefficient of each x^j will be the same as in m(a) times k^(n–j), which is divisible by k unless j = n, which is enough to make the coefficient an integer.
The gibberish in the above paragraph shows that for every algebraic number a, there exists an integer k such that ak is an algebraic integer. Clearly, Q(a) = Q(ak), so every number field can be written as Q(b) where b is in B.
Finally, we can try embedding K into C in various ways. There’s always more than one way – for example, Q(SQRT(2)) can be embedded naturally by sending SQRT(2) to SQRT(2) in C, but it can also be embedded by sending SQRT(2) to -SQRT(2). Algebraically, the resulting field will look exactly the same, because SQRT(2) and -SQRT(2) satisfy the same minimal polynomial, x^2 – 2 = 0. More rigorously, you can check that the function from K to C that sends a + bSQRT(2) to a – bSQRT(2) satisfies the conditions f(cd) = f(c)f(d) and f(c+d) = f(c) + f(d), so we can identify each c in K with f(c) in C. We call any f that satisfies these conditions a ring homomorphism.
If K = Q(a), and m(a) is of degree n, then m(a) has n distinct roots, each of which induces a different ring homomorphism from K to C. To see why, first note that m(a) can be written as the product of linear polynomials over some suitably large field – C is the most obvious, but A is enough. In C[x], m(a)(x) has a as a root, so it’s equal to x–a times a smaller polynomial. That polynomial has a root in C, so we can factor it out as well, and so on. C[x] is a UFD, so the factorization is unique.
Now, if b is a repeated root of m(a) – that is, if m(a) has more than one copy of x–b in its factorization – then it’s also a root of m(a)'(x). The prime is simply a derivative; if m(a)(x) = x^4 + 1, then m(a)'(x) = 4x^3. Derivatives satisfy something called the product rule: (fg)’ = fg‘ + f‘g. So if h^2 divides m(a), then h must divide m(a)’. That means that m(a) and m(a)’ both have b as a root. But then m(b) has to divide both m(a) and m(a)’; but if it divides m(a), it’s either m(a) or a unit, since m(a) is irreducible. It can’t be a unit, so it must be m(a), which doesn’t divide m(a)’. So we have a contradiction.
The above gibberish has just one take-home lesson: m(a) has n distinct roots. So there are n different ring homomorphisms from K to C.
For every b in K, we can then define N(b) to be the product of all of these homomorphisms. When K = Q(SQRT(m)), it turns out to be N(a + bSQRT(m)) = a^2 – mb^2. It can be shown in general that N(b) is always a rational number, and, if b is an algebraic integer, that it’s a rational integer.