Lemma 15.125.9. Let $(R, \mathfrak m)$ be a Noetherian local ring of dimension $d$, and let $f \in \mathfrak m$ be an element not contained in any minimal prime ideal of $R$. Then there exist elements $g_1, \ldots , g_{d - 1} \in \mathfrak m$ and $N \in \mathbf{N}$ such that
$f, g_1, \ldots , g_{d - 1}$ form a system of parameters for $R$
If $h \in \mathfrak m^ N$, then $f + h, g_1, \ldots , g_{d - 1}$ is a system of parameters and we have $\text{length}_ R R/(f, g_1, \ldots , g_{d-1}) = \text{length}_ R R/(f + h, g_1, \ldots , g_{d-1})$.
Proof.
By Lemma 15.125.3 there exist $g_1, \ldots , g_{d - 1} \in \mathfrak m$ such that $f, g_1, \ldots , g_{d - 1}$ is a system of parameters for $R$. Then $\mathfrak m = \sqrt{(f, g_1, \ldots , g_{d - 1})}$. Thus there exists an $n$ such that $\mathfrak m^ n \subset (f, g)$, see Algebra, Lemma 10.32.5. We claim that $N = n + 1$ works. Namely, let $h \in \mathfrak m^ N$. By our choice of $N$ we can write $h = af + \sum b_ ig_ i$ with $a, b_ i \in \mathfrak m$. Thus
\begin{align*} (f + h, g_1, \ldots , g_{d - 1}) & = (f + af + \sum b_ ig_ i, g_1, \ldots , g_{d - 1}) \\ & = ((1 + a)f, g_1, \ldots , g_{d - 1}) \\ & = (f, g_1, \ldots , g_{d - 1}) \end{align*}
because $1 + a$ is a unit in $R$. This proves the equality of lengths and the fact that $f + h, g_1, \ldots , g_{d - 1}$ is a system of parameters.
$\square$
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