Generating a regular sequence out of two 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 $i \geq 2.$ 
 A: 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.
