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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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    Dear @Alex, The definition of a regular ring includes the condition of Noetherianness.2014-08-19

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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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    I think when you prove $A/(x)$ has dim d-1, your actually assume $A/(x)$ is regular local ring, otherwise you can not make sure dim$A/(x)$ is equal to the number of minimal generators.2016-02-20