Krull height theorem

If an ideal of a Noetherian ring is generated by $n$ elements, every prime ideal minimal over it has height at most $n$. The proof inducts on $n$, using the Krull principal ideal theorem in a quotient and prime avoidance to choose a prime chain compatible with the induction.

If $I$ is an $\mathfrak{m}$-primary ideal generated by $n>0$ elements in a Noetherian local ring and $L=\mathrm{length}(R/I)$, then $I^j/I^{j+1}$ is a quotient of $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $R/I$. Summing their lengths gives the displayed bound. Since $I\subseteq \mathfrak{m}$, it also bounds $\mathrm{length}(R/{\mathfrak{m}}^t)$. Together with the prime-chain lower bound for local length, localization at a minimal prime proves the Krull height theorem.