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Let $R$ be a commutative Noetherian ring and $M$ be a nontrivial finitely generated $R$-module. Suppose $(b,a_2,\ldots,a_n)$ and $(c,a_2,\ldots,a_n)$ are $M$-sequences. The following fact is given: if $m_1,\ldots,m_n,m_1',\ldots,m_n'\in M$ are such that

$$ bcm_1+a_2m_2+\cdots+a_nm_n=bm_1'+a_2m_2'+\cdots+a_nm_n' $$

then $m_1'\in(c,a_2,\ldots,a_n)M$.

How to use this fact to deduce that $(bc,a_2,\ldots,a_n)$ is a $M$-sequence?

This problem is from the book Steps in Commutative Algebra by R.Y. Sharp (p. 313).

user26857
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happyEddie
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  • Your equation says $b(cm_1-m_1')\in (a_2,a_3,\ldots,a_n)M$. Now can you finish the proof? – Mohan Oct 07 '15 at 20:01

1 Answers1

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It's easy to see that $bc$ is $M$-regular.
Now let's continue and show that $a_n$ is regular on $M/(bc,a_2,\dots,a_{n-1})M$. Suppose $a_nz\in(bc,a_2,\dots,a_{n-1})M$. Then we can write $$a_nz=bcx_1+a_2x_2+\cdots+a_{n-1}x_{n-1}.$$ But we know that $b,a_2,\dots,a_n$ is an $M$-sequence, so $z\in(b,a_2,\dots,a_{n-1})M$. Write $$z=by_1+a_2y_2+\cdots+a_{n-1}y_{n-1},$$ and plug this in the first equation: $$bcx_1+a_2x_2+\cdots+a_{n-1}x_{n-1}=b(a_ny_1)+a_2(a_ny_2)+\cdots+a_{n-1}(a_ny_{n-1}).$$ Now use what you already know and get $a_ny_1\in(c,a_2,\dots,a_{n-1})M$. Since $c,a_2,\dots,a_{n}$ is an $M$-sequence we obtain $y_1\in(c,a_2,\dots,a_{n-1})M$. Now from $z=by_1+a_2y_2+\cdots+a_{n-1}y_{n-1}$ conclude that $z\in(bc,a_2,\dots,a_{n-1})M$.

user26857
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  • Thank you. I think we also have to show that $a_2$ is not a zero-divisor on $M/(bc)M$, ..., $a_{n-1}$ is not a zero-divisor on $M/(bc,\ldots,a_{n-2})M$ but I suppose the proof is very much similar than for $a_n$. – happyEddie Oct 07 '15 at 20:28
  • @happyEddie I think I should have written $i$ instead of $n$, but you got my point. – user26857 Oct 07 '15 at 20:31