Regular sequences don’t necessarily behave well with respect to permutation or localization under additional hypotheses. However, in all cases they behave well with respect to taking powers. The upshot of this is that the invariant called depth that we will soon introduce is invariant under passing to the radical. We shall deduce this from the following easy fact.

Lemma 4 Suppose we have an exact sequence of {R}-modules

\displaystyle  0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0.

Suppose the sequence {x_1, \dots, x_n \in R} is {M'}-regular and {M''}-regular. Then it is {M}-regular.

The converse is not true, of course.

Proof: Morally, this is the snake lemma. For instance, the fact that multiplication by {x_1} is injective on {M', M''} implies by the snake diagram that {M \stackrel{x_1}{\rightarrow} M} is injective. However, we don’t a priori know that a simple inductive argument on {n} will work to prove this. The reason is that it needs to be seen that quotienting each term by {(x_1, \dots, x_{n-1})} will preserve exactness. However, a general fact will tell us that this is indeed the case. See below. Anyway, this general fact now lets us induct on {n}. If we assume that {x_1, \dots, x_{n-1}} is {M}-regular, we need only prove that {x_{n}: M/(x_1, \dots, x_{n-1})M \rightarrow M/(x_1, \dots, x_{n-1})} is injective. (It is not surjective or the sequence would not be {M''}-regular.) But we have the exact sequence by the next lemma,

\displaystyle  0 \rightarrow M'/(x_1 \dots x_{n-1})M' \rightarrow M/(x_1 \dots x_{n-1})M \rightarrow M''/(x_1 \dots x_{n-1})M'' \rightarrow 0

and the injectivity of {x_n} on the two ends implies it at the middle by the snake lemma.

So we need to prove:

Lemma 5 Suppose {0 \rightarrow M' \rightarrow M \rightarrow M' \rightarrow 0} is a short exact sequence. Let {x_1, \dots, x_m} be an {M''}-sequence. Then the sequence

\displaystyle  0 \rightarrow M'/(x_1 \dots x_m)M' \rightarrow M/(x_1 \dots x_m)M \rightarrow M''/(x_1 \dots x_m)M'' \rightarrow 0

is exact as well.

One argument here uses the fact that the Tor functors vanish when you have a regular sequence like this. We can give a direct argument. It is really just pure diagram-chasing though…

Proof: By induction, this needs only be proved when {m=1}, since we have the recursive description of regular sequences: in general, {x_2 \dots x_m} will be regular on {M''/x_1 M''}. In any case, we have exactness except possibly at the left as the tensor product is right-exact. So let {m' \in M'}; suppose {m'} maps to a multiple of {x_1} in {M}. We need to show that {m'} is a multiple of {x_1} in {M'}. Suppose {m'} maps to {x_1 m}. Then {x_1m} maps to zero in {M''}, so by regularity {m} maps to zero in {M''}. Thus {m} comes from something, {\overline{m}'}, in {M'}. In particular {m' - x_1 \overline{m}'} maps to zero in {M}, so it is zero in {M'}. Thus indeed {m'} is a multiple of {x_1} in {M'}.

So here is the result:

Proposition 6 Let {M} be an {R}-module and {x_1, \dots, x_n} an {M}-sequence. Then {x_1^{a_1} ,\dots, x_n^{a_n}} is an {M}-sequence for any {a_1, \dots, a_n \in \mathbb{Z}_{>0}}.

Proof: The clearest way I know to see this is to use the following lemma, which tells you that regular sequences are stable under certain reasonable actions.

Lemma 7 Suppose {x_1, \dots, x_i, \dots, x_n} and {x_1, \dots, x_i', \dots, x_n} are {M}-sequences for some {M}. Then so is {x_1, \dots, x_i x_i', \dots, x_n}.

Proof: As usual, we can mod out by {(x_1 \dots x_{i-1})} and thus assume that {i=1}. We have to show that if {x_1, \dots, x_n} and {x_1', \dots, x_n} are {M}-sequences, then so is {x_1 x_1', \dots, x_n}. We have an exact sequence

\displaystyle  0 \rightarrow x_1 M/x_1 x_1' M \rightarrow M/x_1 x_1' M \rightarrow M/x_1 M \rightarrow 0.

Now {x_2, \dots, x_n} is regular on the last term by assumption, and also on the first term, which is isomorphic to {M/x_1' M} as {x_1} acts as a nonzerodivisor on {M}. So {x_2, \dots, x_n} is regular on both ends, and thus in the middle. This means that

\displaystyle  x_1 x_1', \dots, x_n

is {M}-regular. That proves the lemma.

So we now can prove the proposition. It is trivial if {\sum a_i = n} (i.e. if all are {1}) it is clear. In general, we can use complete induction on {\sum a_i}. Suppose we know the result for smaller values of {\sum a_i}. We can assume that some {a_j >1}. Then the sequence

\displaystyle  x_1^{a_1}, \dots x_j^{a_j} , \dots x_n^{a_n}

is obtained from the sequences

\displaystyle  x_1^{a_1}, \dots,x_j^{a_j - 1}, \dots, x_n^{a_n}

and

\displaystyle  x_1^{a_1}, \dots,x_j^{1}, \dots, x_n^{a_n}

by multiplying the middle terms. But the complete induction hypothesis implies that both those two sequences are {M}-regular, so we can apply the lemma.

In general, the product of two regular sequences is not a regular sequence. For instance, consider a regular sequence {x,y} in some f.g. module {M} over a noetherian local ring. Then {y,x} is regular, but the product sequence {xy, xy} is never regular.

Advertisements