Let $A$ be a noetherian commutative ring and $\mathfrak{m}$ a maximal ideal. Is it possible to have $ht(\mathfrak{m}) > 1$? Here $ht(\mathfrak{m})$ is the height of $\mathfrak{m}$, that is the supremum of length of chains of prime ideals contained in $\mathfrak{m}$.
I think no but I'm unable to prove it. But my intuition is only derived from examples on finitely generated rings over a field…
EDIT : In fact I was looking for this question : is it possible to have $dim(A/\mathfrak{m}) > 1$? Sorry for the misunderstood… In fact my question comes from a scheme context (I'd like to show that a locally closed point x in a locally noetherian scheme X verify $\bar{\{x\}}$ is of dimension $\leq 1$). I translated it in a commutative algebra fact but I make a mistake in the first formulation.
Best Answer
Consider in $k[x_1,x_2]$ the chain of prime ideals $(0)\subseteq(x_1)\subseteq(x_2,x_2)=:\mathfrak m$.