Lemma 42.68.42. Let $R$ be a local Noetherian ring. Let $\mathfrak q \subset R$ be a prime ideal. Let $M$ be a finite $R$-module such that $\mathfrak q$ is one of the minimal primes of the support of $M$. Then there exist $x_1, \ldots , x_ l \in M$ such that

1. the support of $M / \langle x_1, \ldots , x_ l\rangle$ does not contain $\mathfrak q$, and

2. $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$.

Moreover, in this case $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$.

Proof. The condition that $\mathfrak q$ is a minimal prime in the support of $M$ implies that $l = \text{length}_{R_\mathfrak q}(M_\mathfrak q)$ is finite (see Algebra, Lemma 10.62.3). Hence we can find $y_1, \ldots , y_ l \in M_{\mathfrak q}$ such that $\langle y_1, \ldots , y_ i\rangle / \langle y_1, \ldots , y_{i - 1}\rangle \cong \kappa (\mathfrak q)$ for $i = 1, \ldots , l$. We can find $f_ i \in R$, $f_ i \not\in \mathfrak q$ such that $f_ i y_ i$ is the image of some element $z_ i \in M$. Moreover, as $R$ is Noetherian we can write $\mathfrak q = (g_1, \ldots , g_ t)$ for some $g_ j \in R$. By assumption $g_ j y_ i \in \langle y_1, \ldots , y_{i - 1} \rangle$ inside the module $M_{\mathfrak q}$. By our choice of $z_ i$ we can find some further elements $f_{ji} \in R$, $f_{ij} \not\in \mathfrak q$ such that $f_{ij} g_ j z_ i \in \langle z_1, \ldots , z_{i - 1} \rangle$ (equality in the module $M$). The lemma follows by taking

$x_1 = f_{11}f_{12}\ldots f_{1t}z_1, \quad x_2 = f_{11}f_{12}\ldots f_{1t}f_{21}f_{22}\ldots f_{2t}z_2,$

and so on. Namely, since all the elements $f_ i, f_{ij}$ are invertible in $R_{\mathfrak q}$ we still have that $R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_ i / R_{\mathfrak q}x_1 + \ldots + R_{\mathfrak q}x_{i - 1} \cong \kappa (\mathfrak q)$ for $i = 1, \ldots , l$. By construction, $\mathfrak q x_ i \in \langle x_1, \ldots , x_{i - 1}\rangle$. Thus $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle$ is an $R$-module generated by one element, annihilated $\mathfrak q$ such that localizing at $\mathfrak q$ gives a $q$-dimensional vector space over $\kappa (\mathfrak q)$. Hence it is isomorphic to $R/\mathfrak q$. $\square$

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