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Here is the last problem of my final exam in "Commutative algebra" which I think, no one has solved it completely, today!

Let $R$ be a commutative Noetherian ring. Let $a_1,\dots,a_n$ and $b_1,\dots,b_n$ be two regular sequences in $R.$ Prove that there is a regular sequence $c_1,...,c_n$ s.t. for each $i$, $1 \leq i \leq n$, $$c_i \in (a_1,\dots,a_i) \cap (b_1,\dots,b_i).$$

Note: I attempted to show that $c_i=a_ib_i$ is the desired one, but it seems that we can not do anything, when $i \geq 2.$

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Your attempt does not work for the sequences $(x,y)$ and $(y,x)$ in $k[x,y]$. – Mariano Suárez-Alvarez Jun 20 '11 at 11:46
up vote 5 down vote accepted

Did you learn about the notion of depth in your class? Let $I$ be an ideal in $R$, a commutative noetherian ring. We call a regular sequence $c_1, \dots, c_k \in I$ a maximal $I$-sequence if we cannot find any $i \in I$ such that $c_1, \dots, c_k,i$ is still regular. It is a surprising fact that the length of any maximal $I$-sequence is the same. This length is called the depth of $I$.

It's not at all obvious from first principles that you can define depth, but once it's well-defined your problem becomes much easier:

We do induction: the case $n=1$ is just what you said above. Now take two regular sequences $a_1, \dots, a_n$ and $b_1, \dots, b_n$. Inductively we have that there exists $c_1, \dots, c_{n-1}$ regular such that $c_i \in (a_1, \dots, a_i) \cap (b_1, \dots ,b_i)$. To finish the proof we just need to find $c_n \in (a_1, \dots, a_n) \cap (b_1, \dots ,b_n)$ that is not a zerodivisor mod $(c_1, \dots ,c_{n-1})$. The depth of both $(b_1, \dots ,b_n)$ and $(a_1, \dots, a_n)$ are $n$. Why?

Thus $c_1, \dots, c_{n-1}$ is not a maximal $(\{a_k\})$-sequence or $(\{b_k\})$-sequence. So we can extend it and both: we can find $d \in (\{a_k\})$ and $d' \in (\{b_k\})$ neither $d$ or $d'$ are zerodivisors mod $(c_1, \dots, c_{k-1})$. Thus $dd'$ is not a zero divisor either, but is in both ideals, so it's the desired element.

Bottom line: the notion of depth is really handy. Unfortunately, I don't know a really elementary proof that it's well defined (that all maximal $I$-sequences have the same length). Most proofs use cohomology-- we define a chain complex associated to the generators of the ideal, and the depth is the first place where the complex is not exact.

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Excellent: +1. To the OP: I highly recommend that you accept this answer – zcn Jun 27 '14 at 17:30
I just looked in Kaplansky, and he does give an elementary proof of the well-definedness of depth. You could plug that in here, and make my answer entirely elementary, but it would just look unmotivated and harder. My philosophy on regular sequences is that its always easiest to use the Koszul complex when you can. – Phil Jun 27 '14 at 21:44

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