$(A, \mathfrak{m})$ a Noetherian local ring, $a\in\mathfrak{m}$ a *zero divisor*. Then is it true that $\mbox{dim}\ A/(a) = \mbox{dim}\ A$ ?



Converse of Principal Ideal Theorem
No. Let A be the ring k[x,y,z]/(xz,yz) localized at (x,y,z). Then the dimension of A is 2, but the dimension of A/(x) is one. Note that xz = 0 in A.


The geometric picture is this: In 3-space, take the union of the horizontal z=0 plane with the vertical line x = y = 0, and look at the local ring at the origin. Then restricting to x=0, we get the union of the y-axis and the z-axis.


