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 4Suppose we have an exact sequence of -modules

Suppose the sequence is -regular and -regular. Then it is -regular.

The converse is not true, of course.

*Proof:* Morally, this is the snake lemma. For instance, the fact that multiplication by is injective on implies by the snake diagram that is injective. However, we don’t a priori know that a simple inductive argument on will work to prove this. The reason is that it needs to be seen that quotienting each term by 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 . If we assume that is -regular, we need only prove that is injective. (It is not surjective or the sequence would not be -regular.) But we have the exact sequence by the next lemma,

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

So we need to prove:

Lemma 5Suppose is a short exact sequence. Let be an -sequence. Then the sequence

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 , since we have the recursive description of regular sequences: in general, will be regular on . In any case, we have exactness except possibly at the left as the tensor product is right-exact. So let ; suppose maps to a multiple of in . We need to show that is a multiple of in . Suppose maps to . Then maps to zero in , so by regularity maps to zero in . Thus comes from something, , in . In particular maps to zero in , so it is zero in . Thus indeed is a multiple of in .

So here is the result:

Proposition 6Let be an -module and an -sequence. Then is an -sequence for any .

*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 7Suppose and are -sequences for some . Then so is .

*Proof:* As usual, we can mod out by and thus assume that . We have to show that if and are -sequences, then so is . We have an exact sequence

Now is regular on the last term by assumption, and also on the first term, which is isomorphic to as acts as a nonzerodivisor on . So is regular on both ends, and thus in the middle. This means that

is -regular. That proves the lemma.

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

is obtained from the sequences

and

by multiplying the middle terms. But the complete induction hypothesis implies that both those two sequences are -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 in some f.g. module over a noetherian local ring. Then is regular, but the product sequence is *never* regular.

January 4, 2011 at 3:35 pm

Is it easy to see that Tor_1 vanishes on when we have an M Sequence?

Let x=x1,…,xn be an M-sequence, then is it easy to see that Tor_1(M,R/x)=0?

January 6, 2011 at 12:28 pm

Dear Zac, I am sorry for the slow reply. I suspect that this is not true but could not find a counterexample. You could try asking on math.SE.

January 7, 2011 at 3:46 pm

Hi Akhil,

I thinkthe statementis true but i dont have an easy proof. I picked up this statement from the first line below your Lemma 5. I do feel this statement is correct.