Wednesday, October 17, 2007

Rings and ideals

It's time to do some more algebraic number theory again. For a refresher on what's gone on so far, check here.

In the last installment, way back in June, I introduced rings of algebraic integers, which are the main object of study in algebraic number theory. A "ring" in abstract algebra is a very fundamental concept, first discussed in this article. And in this article the most elementary example of a ring – the rational integers (ℤ) – was discussed, along with the concept of modular arithmetic.

What we saw there was the construction, for any n∈ℤ, of a new ring (ℤ/nℤ), which has only finitely many elements. Modular arithmetic is a staple of elementary number theory (that is, the classical theory of numbers, which deals mainly with ℤ). It was introduced by Carl Friedrich Gauss over 200 years ago, in 1801.

Now we are going to see how that construction can be generalized in abstract ring theory, using the concept of "ideals". It will turn out that the set nℤ consisting of all integer multiples of any n∈ℤ is an example of an ideal, and the "quotient ring" ℤ/nℤ can be generalized for any abstract ring and any ideal of that ring. This construction occurs ubiquitously in the study of rings of algebraic integers, introduced in the last installment.

An ideal can be defined for an arbitrary ring R, but it's a little messy. If R isn't commutative, there can be ideals which are "right" ideals but not "left" ideals or vice versa, because the definition of an ideal involves multiplication. So we'll assume R is commutative and has a multiplicative identity element ("1") too. For such a ring R, consider a subset I⊆R. Then I is an ideal just in case:

  • I is closed under addition: a+b∈I for all a,b∈I.
  • I is closed under multiplication by any element of R:
    ar=ra∈I for all a∈I and r∈R.

These axioms imply that I is a subgroup of R under addition. The additive identity 0 is in I by the second axiom. The second axiom also means that additive inverses are in I because R has a multiplicative identity, and hence its additive inverse -1∈R, so -a∈I for all a∈I. Note that if I≠R, I isn't a full ring (so it isn't a subring of R), because if 1∈I we would have I=R, by the second axiom.

One of the motivations for this concept of ideals is that it makes possible the definition of another very important concept: quotient rings. If as above R is a ring and I is an ideal, the quotient ring, denoted by R/I, is defined as the set of distinct cosets of the form r+I for all r∈R, where r+I is defined for any r∈ R as the set {r+a | a∈I}. Not all cosets are distinct as r ranges over R. Two cosets r+I and r′+I are the same exactly when r-r′∈I. This is because we can write r+I = r′+(r-r′)+I = r′+I. (Because r+I = I if r∈I.)

Importantly, we can define a ring structure on the set of all cosets. Addition is simple: (r+I)+(r′+I) = (r+r′)+I. Multiplication is a little trickier: (r+I)(r′+I) = rr′+I, but this only works since it doesn't depend on the choice of representative of each coset. The problem is we can have r+I=r′′+I even though r≠r′&prime, if r-r′′∈I. But in that case, (r-r′′)r′∈I by the second axiom for ideals, so all is OK. Thus multiplication of cosets is well-defined and unambiguous.

Under these definitions of addition and multiplication R/I is a (commutative) ring with a multiplicative identity. The additive identity element is I, and the multiplicative identity is the coset 1+I.

The ideal structure of the rational integers ℤ provides some important examples. Let n∈ℤ. Then obviously the set nℤ = {nm | m∈ℤ} is an ideal, often written simply as (n). It is just all integral multiples of n. The quotient ring ℤ/nℤ=ℤ/(n) is known as the ring of integers modulo n, and it has n elements. It is familiar from elementary number theory, where one writes "equations" such as a≡b (mod n) just in case a-b is divisible by n, i. e. is a multiple of n, i. e. a-b∈(n). The study of such congruences, which is done all the time in number theory, is really just the study of the ring ℤ/nℤ.

A very important special case is when n is a prime p. Then it is a fact that ℤ/pℤ is a field -- a finite field of p elements, sometimes denoted by Fp. However, if n is not prime, then ℤ/nℤ isn't even an integral domain, because it has divisors of zero, i. e. nonzero elements whose product is 0. For instance, if n=st for s,t∈ℤ, but s,t≠±1, then as ideals (s)≠0 and (t)≠0, where 0=(0)=(n). Yet (s)(t)=0. We can in fact characterize prime numbers p as elements of ℤ such that ℤ/pℤ is a field.

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