A ring of left global dimension zero is a ring $R$ for which every left $R$-module is projective. These are also known as semisimple rings of the Wedderburn-Artin theory fame, which says that these rings are precisely the finite direct products of full matrix rings over division rings. Note the subtle detail that "semisimple" is used here instead of "left semisimple" because left semisimple is the same thing as right semsimple.
In the commutative world, the story for Krull dimension zero is not so simple. For example, every finite commutative ring has Krull dimension zero. Indeed, if $R$ is a ring with Krull dimension greater than zero, then there would exist two distinct primes $P\subset Q$ so that $R/P$ is an integral domain that is not a field. Thus, $R$ is infinite, as every finite integral domain is a field.
The story becomes simpler if we require $R$ to have no nilpotent elements: i.e., that $R$ is reduced. In this case, a commutative ring is reduced and of Krull dimension zero if and only if every principal ideal is idempotent. Every principal ideal being idempotent means that for every $x\in R$ there is an $a\in R$ such that $xax = x$. Rings, commutative or not, satisfying this latter condition are called von Neumann regular. So:
A von Neumann regular ring $R$ must be reduced, because if $x\in R$ is nilpotent then $x^n = 0$ for some minimal $n$, but the existence of an $a\in R$ such that $xax = x$ contradicts the minimality of $n$. An example a non-reduced ring of Krull dimension zero is $\Z/2[t]/t^2$.
It so happens that von Neumann regular rings are exactly the rings for which all left modules and all right modules over them are flat. So, the semsimple rings of the first paragraph are examples of von Neumann regular rings, and it may be amusing to the reader to try prove that a matrix ring over a field is von Neumman regular directly.