I’m just going to start MaBloWriMo here. Anyway, as Qiaochu observes in the comments, National Blog Writing Month (as opposed to National Novel Writing Month) was in October. Before I start, I think it might be appropriate to post my half-complete notes on commutative algebra from a course I’m taking. Since I’m not going to be self-contained, they might be a helpful reference for readers.

For the next few posts, we shall always assume that all rings are commutative and noetherian. Commutativity is always needed. Noetherianness won’t be needed for now, but it will soon become indispensable. So let {R} be a ring, and {M} an {R}-module. We want to talk about the definition of a regular sequence on {M}. This is going to be a sequence of elements of {R} that act “independently” on {M} in some sense. We are going to make this precise below when we interpret a consequence of this condition via associated graded modules. This definition is basically the groundwork for everything that follows, as you need it for the definitions of notions such as Cohen-Macaulayness and regularity. Let us start by stating the definition.

Definition 1 A sequence {x_1, \dots, x_n \in M} is {M}-regular (or is an {M}-sequence if for each {k \leq n}, {x_k} is a nonzerodivisor on the {R}-module {M/(x_1, \dots, x_{k-1}) M} and also {(x_1, \dots, x_n) M \neq M}.

So {x_1} is a nonzerodivisor on {M}, by the first part. That is, the homothety {M \stackrel{x_1}{\rightarrow} M} is injective. The last condition is also going to turn out to be necessary for us.

Example 1 The basic example one is supposed to keep in mind is the polynomial ring {R = R_0[x_1, \dots, x_n]} and {M = R}. Then the sequence {x_1, \dots, x_n} is regular in {R}.

Suppose now that {M} is finitely generated. Things become the nicest in this context. Then it is a basic fact of commutative algebra that if {N} is any {R}-module, the guys in {R} that are zerodivisors on {N} are precisely those that lie in the union of the associated primes of {N}. This fact will become crucial to us as we attempt to construct regular sequences. The property of being a regular sequence is inherently an inductive one. Note that {x_1, \dots, x_n} is a regular sequence on {M} if and only if {x_1} is a zerodivisor on {M} and {x_2, \dots, x_n} is an {M/x_1 M}-sequence.

0.1. Basic properties

The first observation to make is that regular sequences are not preserved by permutation. This is one nice characteristic that we would like but is not satisfied. Nonetheless,

Proposition 2 Let {R} be a noetherian local ring and {M} a finite {R}-module. Then if {x_1, \dots, x_n} is a {M}-sequence contained in the maximal ideal, so is any permutation {x_{\sigma(1)}, \dots, x_{\sigma(n)}}.

Proof: It is clearly enough to check this for a transposition. Namely, if we have an {M}-sequence

\displaystyle  x_1, \dots, x_i, x_{i+1}, \dots x_n

we would like to check that so is

\displaystyle  x_1, \dots, x_{i+1}, x_i, \dots, x_n.

It is here that we use the inductive nature. Namely, all we need to do is check that

\displaystyle  x_{i+1}, x_i, \dots,x_n

is regular on {M/(x_1, \dots,x_{i-1}) M}, since the first part of the sequence will automatically be regular. Now {x_{i+2}, \dots, x_n} will automatically be regular on {M/(x_1, \dots, x_{i+1})M}. So all we need to show is that {x_{i+1}, x_i} is regular on {M/(x_1, \dots, x_{i-1})M}. The moral of the story is that we have reduced to the following lemma. Let {N} be a finite {R}-module and {a,b \in R} an {N}-sequence contained in the maximal ideal. Then so is {b,a}. We can prove this as follows. First, {a} will be a nonzerodivisor on {N/bN}. Indeed, if not then we can write

\displaystyle  an = bn'

for some {n,n' \in N} with {n \notin bN}. But {b} is a nonzerodivisor on {N/aN}, which means that {bn' \in aN} implies {n' \in aN}. Say {n' = an''}. So {an = ba n''}. As {a} is a nonzerodivisor on {N}, we see that {n = bn''}. Thus {n \in bN}, contradiction. This part has not used the fact that {R} is local. Now I claim that {b} is a nonzerodivisor on {N}. Suppose {n \in N} and {bn = 0}. Since {b} is a nonzerodivisor on {N/aN}, we have that {n \in aN}, say {n = an'}. Thus

\displaystyle  b(an') = a(bn') = 0.

The fact that {N \stackrel{a}{\rightarrow} N} is injective implies that {bn' = 0}. So we can do the same and get {n' = an''}, {n'' = a n^{(3)}, n^{(3)} =a n^{(4)}}, and so on. It follows that {n} is a multiple of {a, a^2,a^3, \dots}, and hence in {\mathfrak{m}^j N} for each {j} where {\mathfrak{m} \subset R} is the maximal ideal. The Krull intersection theorem now implies that {n = 0}. Together, these arguments imply that {b,a} is an {N}-sequence, proving the lemma.

One might wonder what goes wrong; after all, oftentimes we can reduce results to their analogs for local rings. Yet the fact that regularity is preserved by permutations for local rings does not extend to arbitrary rings. The problem is that regular sequences do not localize. Well, they almost do, but the final condition that {(x_1, \dots, x_n) M \neq M} doesn’t get preserved. We can state:

Proposition 3 Suppose {x_1, \dots, x_n} is an {M}-sequence. Let {N} be a flat {R}-module. Then if {(x_1, \dots, x_n)M \otimes N \neq M \otimes N}, then {x_1, \dots, x_n} is an {M \otimes N}-sequence.

Proof: This is actually very easy now. The fact that {x_i: M/(x_1, \dots, x_{i-1})M \rightarrow M/(x_1, \dots, x_{i-1})M} is injective is preserved when {M} is replaced by {M \otimes N} because the functor {- \otimes N} is exact.

In particular, it follows that if we have a good reason for supposing that {(x_1,\dots, x_n) M \otimes N \neq M \otimes N}, then we’ll already be done. For instance, if {N} is the localization of {R} at a prime ideal containing the {x_i}. Then we see that automatically {x_1, \dots, x_n} is an {M_{\mathfrak{p}} = M \otimes_R R_{\mathfrak{p}}}-sequence.

Next time, we’ll continue talking about the fundamentals of what this property really means.