See and [Lemma 2 page 300, MatCA].

Lemma 51.17.3. Let $A$ be a ring. If $f_1, \ldots , f_{r - 1}, f_ rg_ r$ are independent and if the $A$-module $A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)$ has finite length, then

\begin{align*} & \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ rg_ r)) \\ & = \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, f_ r)) + \text{length}_ A(A/(f_1, \ldots , f_{r - 1}, g_ r)) \end{align*}

Proof. We claim there is an exact sequence

$0 \to A/(f_1, \ldots , f_{r - 1}, g_ r) \xrightarrow {f_ r} A/(f_1, \ldots , f_{r - 1}, f_ rg_ r) \to A/(f_1, \ldots , f_{r - 1}, f_ r) \to 0$

Namely, if $a f_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$, then $\sum _{i < r} a_ i f_ i + (a + bg_ r)f_ r = 0$ for some $b, a_ i \in A$. Hence $\sum _{i < r} a_ i g_ r f_ i + (a + bg_ r)g_ rf_ r = 0$ which implies $a + bg_ r \in (f_1, \ldots , f_{r - 1}, f_ rg_ r)$ which means that $a$ maps to zero in $A/(f_1, \ldots , f_{r - 1}, g_ r)$. This proves the claim. To finish use additivity of lengths (Algebra, Lemma 10.51.3). $\square$

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