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Let $(A,\frak m)$ be a local Noetherian ring and let $x_1,\dots,x_d$ be a system of parameters, i.e. ${\frak m}=(x_1,\dots,x_d)$. Then $$\dim A/(x_1,\dots,x_i)=d-i$$ for each $i=1,\dots,d$.

I know just a few basic facts about dimension theory. I think I can prove the inequality $\le$ via Krull's Hauptidealsatz in this way: the maximal ideal of $A/(x_1,\dots,x_i)$ is $$\mathfrak m_i:= (\overline{x_{i+1}},\dots,\overline{x_d}).$$ So it must be $\mathrm{ht}(\mathfrak m_i)\le d-i$.

But how to prove the other inequality? I think I should do it by induction, but I cannot understand how to begin. So, if what I said so far is right (is it?), my question is: how can I prove that $$\dim A/(x_1)\ge d-1?$$

Thank you very much, Miles

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Your proof for the 1st part is not true:
$x_1,\dots,x_d$ being system of parameters does not mean ${\frak m}=(x_1,\dots,x_d),$ So $(\overline{x_{i+1}},\dots,\overline{x_d})$ isn't necessarily the maximal ideal of $A/(x_1,\dots,x_i).$

For complete proof of this, see "PROPOSITION 15.22" of the book "Steps in Commutative Algebra" By "R. Y. Sharp" (pages $297,298$). Search the book by google, if you dont have access to it.

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  • $\begingroup$ I will read it carefully, thank you. $\endgroup$ – Miles Eagle Jul 25 '16 at 9:00
  • $\begingroup$ I have seen now for the first time what you mean. I am sorry for my blindness. $\endgroup$ – Miles Eagle Jul 3 '17 at 10:56
  • $\begingroup$ you are welcome $\endgroup$ – user 1 Jul 3 '17 at 11:06

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