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I'm studying these notes, in particular page 17 theorem 1. I have some problems in the implication iii $\Rightarrow$ ii:

Let $A$ be a local ring of dimension $n$ and maximal ideal $m$

ii) $A$ is Cohen-Macaulay and there exists a system of parameter generating an irreducible ideal (irreducible means I cannot see it as an intersection of 2 different ideals)

iii) every system of parameters generates an irreducible ideal.

Here it is how the proof goes:

Induction on $\mathrm{dim}\;A$. If it is 0 we are ok. Suppose $\mathrm{dim}\;A=n>0$, we want to prove $\mathrm{depth}\;A\geq1$. Suppose $\underline{f}=(f_1,\ldots,f_n)$ is a system of parameters. Define $\underline{f}_r=(f_1^r,\ldots,f_n^r)$, it is another system of parameters. We have

$\mathrm{Hom}_A(k,\underline{f}_r/\underline{f}_{r+1})\neq0$ Why?

then we have:

$\mathrm{Hom}_A(k,A/\underline{f}_{r+1})$ is a $k$-vectorspace of dimension 1, Why?

this should imply

$\underline{f}_{r+1}:m\subset\underline{f}_r$ for every $r$, why?

And so $0:m\subset\bigcap_r(\underline{f}_{r+1}:m)\subset\bigcap_r\underline{f}_r=0$.

And now we can easily conclude.

where did we use the hypothesis that the $\underline{f}_r$ are irreducible?

Could you help me in solving these 4 doubts that I have?

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If you like, I can prove directly that $\underline{f}_{r+1}:m\subset\underline{f}_r$ for every $r$. – user26857 Oct 29 '12 at 2:35
ok let's see how you prove it – Chris Oct 29 '12 at 3:12
up vote 0 down vote accepted

I will show that $\underline{f}_{r+1}:\mathfrak{m}\subset\underline{f}_r$ for every $r\ge 1$. Assume the contrary and let $x\in \underline{f}_{r+1}:\mathfrak{m}$, $x\notin\underline{f}_r$.

Claim. $\underline{f}_{r+1}=(\underline{f}_{r+1}+xA)\cap\underline{f}_r$.

Proof. Pick $a\in(\underline{f}_{r+1}+xA)\cap\underline{f}_r$ and write $a=b+xr$ with $b\in\underline{f}_{r+1}$, $r\in A$. Then $a-b=xr\in\underline{f}_r$ and $\underline{f}_r$ is $\mathfrak{m}$-primary, so $r\in\mathfrak{m}$ and this is enough to show that $a\in\underline{f}_{r+1}$.

Since $\underline{f}_{r+1}$ is irreducible, then $\underline{f}_{r+1}=\underline{f}_{r+1}+xA$ or $\underline{f}_{r+1}=\underline{f}_{r}$. Because $x\notin\underline{f}_r$ and $\underline{f}_{r+1}\subset\underline{f}_{r}$, the first equality is impossible. It remains that $\underline{f}_{r+1}=\underline{f}_{r}$. Since $\underline{f}_{r+1}\subset\underline{f}_{1}\underline{f}_{r}$ we get $\underline{f}_{r+1}\subset\underline{f}_{1}\underline{f}_{r+1}$, that is, $\underline{f}_{r+1}=\underline{f}_{1}\underline{f}_{r}$. It follows (by Nakayama) that $\underline{f}_{r+1}=0$, and this implies that $A$ is artinian, contradiction.

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