The tome Commutative Algebra by Zariski and Samuel opens with the memorable sentence, “This book is the child of an unborn parent.” As Zariski explains,
Some years ago the senior author began the preparation of a Colloquium volume on algebraic geometry, and he was then faced with the difficult task of incorporating in that volume the vast amount of purely algebraic material which is needed in abstract algebraic geometry … Thus the idea of a separate book on commutative algebra was born, and the present book [appears] … when its parent … has still to see the light of the day.
Zariski and Samuel is comprehensive. This post lies at the opposite extreme: I want to sketch the bare minimum of commutative algebra that we’ll need for the next few posts. All very standard stuff, so if rings, ideals, PIDs and UFDs are old friends, skip ahead. Except do glance at the stuff at the end about congruences—also standard, but many authors don’t use this notation as much as I think they should.
I won’t be proving anything, but it’s easy to find treatments online.
OK. Ring will always mean commutative ring with multiplicative identity (aka unity, or 1). (Some people like to write “rng” when the unity might be missing. Cute.) In the so-called zero ring, and only there, 0=1. The zero ring has only that one element (since r=r1=r0=0, for all r).
We have the usual definition of homomorphism φ:R→S (φ(r+s)=φ(r)+φ(s), etc.), which includes the requirement of preserving unity: φ(1)=1. The kernel of φ (i.e., φ–1(0)) is an ideal: a subset I⊆R that is closed under addition and under multiplication by arbitrary elements of R. (Briefly, I+I ⊆ I and RI ⊆ I.) Conversely, given an ideal I⊆R, we can construct the quotient ring R/I, with the canonical homomorphism R→R/I. One description: define r ≡ s mod I (r is congruent to s) if r–s∈I. The elements of R/I are the equivalence (better, congruence) classes. Trivial fact: r+I = {s : s≡r mod I}. You define the operations on R/I in the obvious way, and prove they are well-defined using the fact that I is an ideal. The canonical homomorphism is r↦r+I, and is onto.
If A is a subset of R, the smallest ideal containing A is the set of all finite sums of the form Σ riai with all ai∈A. Special case: the principal ideal (a) = Ra, i.e., all multiples of a. Extra-special cases: the zero ideal (0)={0} and the whole ring R=(1). Note that R/(0) is R, and R/(1) is the zero ring. (Or if you want to be fussy, “isomorphic to”. But it’s better to chill out over this sort of thing.) Another special case we’ll be seeing a lot of: (a,b) = Ra+Rb= all sums of the form ra+sb, r,s∈R.
Two bigshot rings in the mathematical universe: the integers ℤ, and the ring k[x1,…,xn] of all polynomials in n variables over a field k. Both are integral domains, i.e., are nonzero rings having no zero divisors, i.e., nonzero elements r,s satisfying rs=0. You can always cancel in an integral domain: ra = rb implies a = b if r ≠ 0. Any integral domain has an associated fraction field; for ℤ this is ℚ, the rational numbers, and for k[x1,…,xn] it is k(x1,…,xn), the field of rational expressions in the n variables.
Both ℤ and k[x1,…,xn] are unique factorization domains, UFDs for short. This means they are integral domains where every nonzero element can be factored into irreducibles, uniquely up to units. We say u is a unit of the ring if it has an inverse, so uv=1 for some v. We say ur is an associate of r when u is a unit. The units of ℤ are ±1; the units of k[x1,…,xn] are the polynomials of degree 0, i.e., just the elements of k. An element p is irreducible if it is not a unit, and a factorization p=rs implies that r or s is a unit (and so the other factor is an associate). Incidentally, the proof that k[x1,…,xn] is a UFD relies on one of Gauss’s many lemmas.
ℤ and k[x] share an even stronger property: they are principal ideal domains (PIDs), meaning that every ideal is principal. A key consequence: the ideal (a,b)=(c) for some c, so ra+sb = c for some r and s; it follows that c is a gcd of a and b, i.e., a common divisor divisible by all common divisors (since any divisor of a and b divides ra+sb). Easy-to-prove fact: gcds are unique up to units in UFDs. (So instead of saying a gcd, we’ll chill out and say the gcd.) Not-hard-to-prove fact: all PIDs are UFDs. Easy observation: k[x,y] is not a PID, but as we’ve noted, it’s a UFD.
People sometimes call ra+sb = c the Bézout identity (same Bézout!); it will play an important role for us. Here’s another consequence: in a PID, every irreducible element p is prime, meaning that if p divides ab then it divides a or b. (Trick: if p doesn’t divide a, then the gcd of p and a is 1, so rp+sa=1 for some r and s. Now multiply through by b.) If the gcd of a and b is 1, we say a and b are coprime or relatively prime. So in a PID, we have ra+sb =1 for coprime a and b.
Now, all the stuff of the last paragraph occurs in elementary number theory. But if we translate it into the language of ideals, important generalizations fall out. A few observations: a|b is equivalent to b∈(a), and also to (a)⊇(b). Next: p is prime if and only if it is not a unit and ab∈(p) implies a∈(p) or b∈(p). We have (u)=R if and only if u is a unit, and (a)=(b) if and only if a and b are associates. So p is irreducible if and only if (p)≠R and (p)⊂(r)⊂R is impossible, i.e., any divisor r of p is either an associate or a unit. (I use ⊂ for proper containment.)
Definition: an ideal P is prime if ab∈P implies a∈P or b∈P and P≠R. Routine exercise: P is prime if and only if R/P is an integral domain. An ideal M is maximal if M≠R and M⊂I⊂R is impossible, I being an ideal of course. Not-hard exercise: M is maximal if and only if R/M is a field.
I’m getting tired, probably you too. So let me say a little about how this relates to algebraic geometry, and finish off with my pitch for congruence notation.
The ring k[x1,…,xn] takes the starring role. (For us mainly k[x,y] .) Let’s write R for k[x1,…,xn], and A for kn, so-called affine n-space. R and A have a dance to perform, a kind of tango. If I is any ideal in R, and S is any subset of A, define:
- V(I) = all points (a1,…,an) in A such that f(a1,…,an)=0 for all f∈I. In other words, it’s the set of simultaneous solutions to all the equations f(x1,…,xn)=0. In yet other words, it’s the set of points where all the curves f(x1,…,xn)=0 intersect, f∈I. (The old-fashioned term is locus.)
- I(S) = all polynomials f in R such that f(a1,…,an)=0 for all (a1,…,an)∈S. (Easy to see that this is an ideal.)
Sets of the form V(I) are called algebraic sets, or sometimes varieties. (Some people use variety in a more restricted sense.)
It is not the case that the V and I mappings furnish a 1-1 correspondence between subsets and ideals—typically, V(I(S))⊃S and I(V(I))⊃I. If this post were a textbook, I’d immediately hit you with crank-turning exercises galore on these mappings. (Most basic: order reversing with respect to containment.) But I should mention Hilbert’s “big two” theorems in the subject (not routine exercises, of course).
- Hilbert Basis Theorem: any ideal in k[x1,…,xn] is finitely generated, so I=(f1,…,fm) for some polynomials (f1,…,fm).
- Hilbert’s Nullstellensatz (weak form): If k is algebraically closed, then I=(1)=R if and only if V(I)=∅. In other words, if I≠(1), then the equations in I have a simultaneous solution. (The other direction is trivial: 1=0 has no solutions.)
- Hilbert’s Nullstellensatz (strong form): For any ideal I of k[x1,…,xn], I(V(I)) is the so-called radical of I, often denoted
. This is the set {q∈k[x1,…,xn] : for some positive integer m, qm ∈ I}.
The strong form is a corollary to the weak form, using something called Rabinowitsch’s trick (see Fulton pp.10–11).
OK, here’s my spiel for congruence notation. A simple example: let R = k[x,y], and let I be the principal ideal (y–xn) for some n. So y ≡ xn. Lots of consequences just fall right out, e.g., x3y2 ≡ x3+2n. Now, you can rephrase this purely in terms of ideals: y—xn ∈ I, hence both y(y–xn)=y2–yxn ∈ I and (y–xn)xn=yxn–x2n ∈ I, hence their sum y2–x2n ∈ I, hence x3(y2–x2n)=x3y2–x3+2n ∈ I, hence x3y2 and x3+2n belong to the same coset of I. But using congruence notation, you just roll down a hill (so to speak).
If π:R→R/I is the canonical homomorphism, then a ≡ b mod I if and only if π(a) = π(b). We’ll be spending a lot of time with k[x,y]/(E,F) in the next batch of posts.
Diagonal Argument 