Question 1.224: Suppose R is a Noetherian local ring. Prove that if M is the......
Suppose R is a Noetherian local ring. Prove that if M is the unique maximal ideal of R and \operatorname{dim}_{R / M} M / M^{2} is a finite dimensional vector space over R / M. If d=\operatorname{dim}_{R / M} M / M^{2}, then prove that any generating set for M has at least d elements.
Let U denote M / M^{2}. For m \in M and r \in R we denote by \bar{m} and \check{r} the images of m and r in U and R / M, respectively. By the action \check{r} \cdot \bar{m}=\overline{r m} of R / M, U becomes a module over R / M. Since R is Noetherian, every ideal in R is finitely generated. In particular M is finitely generated. The image in U of a generating set for M is a generating set for U as an R-module. In particular, U is finitely generated R-module. But any element from M acts as a zero on U, thus U is a finitely generated R / M module, hence a finite dimensional vector space. Let d=\operatorname{dim}_{R / M} M / M^{2} be the vector space dimension of M / M^{2} and let x_{1}, \ldots, x_{n} be a list of generators of M as an ideal. Since \overline{x_{1}}, \ldots, \overline{x_{n}} in U is a spanning set we see that n \geq d.