Lemma 15.125.8. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$, let $t$ be the number of minimal prime ideals of $R$ of dimension $d$, and let $(g_1,\ldots ,g_ d)$ be a system of parameters. Then $t \leq \text{length}_ R(R/(g_1,\ldots ,g_ n))$.

**Proof.**
If $d = 0$ the lemma is trivial. If $d = 1$ the lemma is Lemma 15.125.2. Thus we may assume $d > 1$. Let $\mathfrak p_1, \ldots , \mathfrak p_ s$ be the minimal prime ideals of $R$ where the first $t$ have dimension $d$, and denote $I = (g_1, \ldots , g_ n)$. Arguing in exactly the same way as in the proof of Lemma 15.125.2 we can assume $R$ is reduced.

Assume $R$ is reduced with minimal primes $\mathfrak p_1, \ldots , \mathfrak p_ t$. This means there is an exact sequence

Here $Q$ is the cokernel of the first map. Write $M = \prod _{i = 1}^ t R/\mathfrak p_ i$. Localizing at $\mathfrak p_ j$ we see that

is surjective. Thus $Q_{\mathfrak p_ j} = 0$ for all $j$. Therefore no height $0$ prime of $R$ is in the support of $Q$. It follows that the degree of the numerical polynomial $n \mapsto \text{length}_ R(Q/I^ nQ)$ equals $\dim (\text{Supp}(Q)) < d$, see Algebra, Lemma 10.62.6. By Algebra, Lemma 10.59.10 (which applies as $R$ does not have finite length) the polynomial

has degree $< d$. Since $M = \prod R/\mathfrak p_ i$ and since $n \to \text{length}_ R(R/\mathfrak p_ i + I^ n)$ is a numerical polynomial of degree exactly(!) $d$ for $i = 1, \ldots , t$ (by Algebra, Lemma 10.62.6) we see that the leading coefficient of $n \mapsto \text{length}_ R(M/I^ nM)$ is at least $t/d!$. Thus we conclude by Lemma 15.125.7. $\square$

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