Lemma 90.3.3. Let $f: B \to A$ be a surjective ring map in $\mathcal{C}_\Lambda $. Then $f$ can be factored as a composition of small extensions.
Proof. Let $I$ be the kernel of $f$. The maximal ideal $\mathfrak {m}_ B$ is nilpotent since $B$ is Artinian, say $\mathfrak {m}_ B^ n = 0$. Hence we get a factorization
\[ B = B/I\mathfrak {m}_ B^{n-1} \to B/I\mathfrak {m}_ B^{n-2} \to \ldots \to B/I \cong A \]
of $f$ into a composition of surjective maps whose kernels are annihilated by the maximal ideal. Thus it suffices to prove the lemma when $f$ itself is such a map, i.e. when $I$ is annihilated by $\mathfrak {m}_ B$. In this case $I$ is a $k$-vector space, which has finite dimension, see Algebra, Lemma 10.53.6. Take a basis $x_1, \ldots , x_ n$ of $I$ as a $k$-vector space to get a factorization
\[ B \to B/(x_1) \to \ldots \to B/(x_1, \ldots , x_ n) \cong A \]
of $f$ into a composition of small extensions. $\square$
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