Dimension theorem for local rings

Remarks

The usual proof shows $\delta(R) \ge d(R) \ge \dim R \ge \delta(R)$. The first inequality holds because the Hilbert–Samuel polynomial of an ideal generated by $s$ elements has degree at most $s$, and its degree does not depend on the $\mathfrak m$-primary ideal. The second goes by induction on $d(R)$. Take a chain of primes $\mathfrak p_0 \subsetneq \cdots \subsetneq \mathfrak p_r$ and an element $x \in \mathfrak p_1 \setminus \mathfrak p_0$. Then the ring $R/(\mathfrak p_0 + (x))$ has a smaller $d$ and still contains a chain of length $r - 1$. The third goes by building, with the prime avoidance lemma, elements $x_1, \dots, x_{\dim R}$ such that each $x_i$ lies outside every prime of height $i - 1$ that contains $x_1, \dots, x_{i-1}$. For example, for $R = k[x, y]_{(x, y)}$ with $k$ a field, $\chi(n) = n(n+1)/2$, of degree $2$, the ideal $\mathfrak m = (x, y)$ needs two generators, and $\dim R = 2$. Consequences: $\dim R \le \dim_k \mathfrak m/\mathfrak m^2$ with $k = R/\mathfrak m$, so $\dim R$ is finite, and Krull's height theorem follows.

Going deeper

For a student

The theorem turns the order-theoretic Krull dimension into something computable. The number $\delta(R)$ gives upper bounds, since any $\mathfrak m$-primary ideal with $s$ generators shows $\dim R \le s$. The degree $d(R)$ connects dimension with Hilbert polynomials and with the multiplicity of the ring. A list of $\dim R$ elements that generates an $\mathfrak m$-primary ideal is a system of parameters.