Krull dimension of a noncommutative ring

Remarks

From clause 2 with $\alpha = 0$, a module has Krull dimension $0$ exactly when it is nonzero and Artinian: the factors of every descending chain are eventually $0$.

Example: $\operatorname{Kdim} \Z_\Z = 1$. Take a descending chain $M_0 \supseteq M_1 \supseteq \cdots$ of subgroups of $\Z$. If every $M_i = a_i\Z$ is nonzero, every factor $M_i / M_{i+1}$ is finite, hence Artinian. Otherwise some $M_i = 0$, and all later factors are $0$. So $\operatorname{Kdim} \Z \le 1$, and $\Z$ is not Artinian, so $\operatorname{Kdim} \Z = 1$.

Non-example: an infinite-dimensional vector space $V$ over a field has no Krull dimension. Splitting a basis into infinitely many infinite parts gives a descending chain $V \supsetneq M_1 \supsetneq M_2 \supsetneq \cdots$ in which every factor is infinite-dimensional. So an induction on $\alpha$ shows that $\operatorname{Kdim} V \le \alpha$ fails for every ordinal $\alpha$.

Every Noetherian module has a Krull dimension. For a commutative Noetherian ring, the Krull dimension defined here equals the classical Krull dimension given by chains of prime ideals. The definition is the deviation of the partially ordered set of submodules of $M$.