Regular local rings have no excess generators: $\dim_k \mfm/\mfm^2=\dim R$.Suggest a slogan
A Noetherian local ring \((R,\mfm)\) with residue field \(k=R/\mfm\) is regular if \(\mfm\) can be generated by \(\dim R\) elements, where \(\dim R\) is the Krull dimension of \(R\). By Nakayama's lemma for a local ring, the minimal number of generators of \(\mfm\) is \(\dim_k\mfm/\mfm^2\), so \(R\) is regular if and only if \(\dim_k\mfm/\mfm^2=\dim R\).
A commutative Noetherian ring \(R\) is regular if the localization \(R_{\mfp}\) is a regular local ring for every prime ideal \(\mfp\subseteq R\).
By Krull’s height theorem, the extension of Krull's Hauptidealsatz to ideals generated by \(r\) elements, \(\dim R\le\dim_k\mfm/\mfm^2\) for every Noetherian local ring \((R,\mfm)\); regularity is the case of equality.
What is a regular ring?