If all prime ideals are maximal, then why is krull dimension 0?

Question

I know that if all prime ideals are maximal ideals (like in PID or Artin ring) then why is their krull dimension $0$? Let's say we have $p$ a prime ideal in Artin ring, then it is maximal so we have $(0)\subsetneq p$ where strict chain of prime ideals $(0)$ and $p$ (which is maximal therefore prime) would give krull dimension of $1$ not $0$ for Artin ring.

Also, for a PID that is not a field, the krull dimension is $1$. Again all nonzero prime ideals are maximal in PID therefore for $(0)\subsetneq p$, where chain of prime ideals $(0)$ and $p$ would give it dimension $1$ as expected.

So is the logic for both PID that's not a field and the Artin ring same for dimension? Then their dimensions should be $1$. I am slightly confused with the definition. Please explain.

Answer 1

Looks like you've fallen into two misapprehensions:

so we have $(0)\subsetneq p$ where $p$ is maximal. Then the krull dimension should be $1$ not $0$.

You did not exhibit a chain of prime ideals. In a right Artinian ring, the zero ideal is only prime if the ring is already a simple ring, and so there aren't any other proper ideals to be had. In particular for a commutative Artinian ring, the zero ideal is prime only if the ring is already a field.

Also, for a PID that is not a field, the krull dimension is $1$. Again all prime ideals are maximal in PID, so $(0)⊊𝑝$ would give it dimension 1 as expected.

All nonzero prime ideals are maximal in a PID, but the ideal $\{0\}$ is prime and not maximal, so a nonfield PID does not satisfy what is required. (What was required was all primes maximal.)

Proving the statement is straightforward enough: suppose you have a prime ideal $P$ that isn't maximal. Pick any maximal ideal $M$ containing $P$. $M$ is also prime. Then $P\subsetneq M$ witnesses that the Krull dimension is at least $1$. Using the contrapositive, primes are maximal in a zero-dimensional ring.