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How can I prove this:

Let $A$ be a local regular ring with maximal ideal $\mathfrak m$ and $x \in \mathfrak m-\mathfrak m^2$. Then $A/(x)$ is a regular ring. Prove also that if $x\in\mathfrak m^2$, $x\ne 0$, the result does not hold anymore.

I don't know how to begin! Thanks to everyone who helps!

But if $d=\infty$?

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If $d=\infty$ then the ring is no longer Noetherian as $m$ isn't finitely generated, so all bets are off. But I'm guessing your definition of "local" requires that the ring be Noetherian. –  Alex Becker Jan 6 '12 at 0:24
    
Dear @Alex, The definition of a regular ring includes the condition of Noetherianness. –  Keenan Kidwell Aug 19 at 20:02

2 Answers 2

up vote 2 down vote accepted

Hint: Let $d$ be the Krull dimension of $A$. Show that $x$ is part of a set $x,x_2,\ldots,x_d\in A$ such that $(x,x_2,\ldots,x_d) = \mathfrak m$ iff $x\in \mathfrak m-\mathfrak m^2$. From this (and using the fact that there is a bijection between the maximal ideals of $A$ containing $(x)$ and the maximal ideals of $A/(x)$), you can see that $A/(x)$ is a local ring with maximal ideal $(x_2,\ldots,x_d)$, which has Krull dimension at most $d-1$ by Krull's Principal Ideal Theorem. The dimension is in fact exactly $d-1$, as if we had fewer than $d-1$ generators for $\mathfrak m/(x)$ then adjoining $x$ gives you fewer than $d$ generators for $\mathfrak m$, violating Krull's PIT. Since $\mathfrak m/(x)$ is generated by $d-1$ elements, this makes $A/(x)$.

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Using this lemma can i directly say that the ring $A=K[x,y,z]_{(x,y,z)}/(x-y^2-z^2)$ is local regular and $B=K[x,y,z]_{(x,y,z)}/(x^2-y^2-z^2)$ is not, right? –  balestrav Jan 5 '12 at 19:27
    
One more doubt. So we can prove that if $x \in m^2$ then $A/(x)$ is not regular, or sometimes we can have $x\in m^2$ and still $A/(x)$ regular? –  balestrav Jan 5 '12 at 20:14
    
@balestrav As per your first comment, you can show that $A$ is local regular using this by showing $x-y^2-z^2\in m-m^2$, but not that $B$ is not. This gets into your second comment. I do not know if there is any counterexample to $x\in m^2\implies A/(x)$ not regular local, but I don't see a proof either. All that the question seems to ask is to prove that $A/(x)$ need not be regular local. This doesn't go against my hint, because the first part of the hint is both a sufficient and necessary condition for existence of a system of parameters, but what follows are just sufficient conditions for –  Alex Becker Jan 5 '12 at 23:36
    
the ring $A/(x)$ to be regular local. However, it should suggest to you some examples to show that $A/(x)$ need not be regular local if $x\in m^2$. Technically, you only need to show "if" for the first part, not "only if". –  Alex Becker Jan 5 '12 at 23:37
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I think i've proven the Krull dimension to be less than 2, while the dimension of $m/m^2$ is 3, so it's not regular.. –  balestrav Jan 6 '12 at 1:50

If $(A,\mathfrak m)$ is a local regular ring and $x\in\mathfrak m^2$, $x\ne 0$, then $A/(x)$ is not regular since $\dim A/(x)=\dim A-1$ (note that $A$ is an integral domain), and $\operatorname{edim} A/(x)=\operatorname{edim} A$, where $\operatorname{edim} A:=\dim_{A/\mathfrak m}\mathfrak m/\mathfrak m^2$ (note that the maximal ideal of $A/(x)$ is $\mathfrak m/(x)$ and its square is $\mathfrak m^2/(x)$, so $\dfrac{\mathfrak m/(x)}{\mathfrak m^2/(x)}=\mathfrak m/\mathfrak m^2$).

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