Fiber product of local Artinian rings with fixed residue field

Question

Let $\Lambda$ be a complete Noetherian local ring with residue field $k$ and suppose $A \to C$ and $B \to C$ are local homomorphisms of Artinian local $\Lambda-$algebras with residue field $k$. The question fiber product of local artinian rings why the fiber product $A \times_C B$ is still Artinian. However, I don't understand the accepted answer. The answer seems to be saying that $A$ being Artinian (i.e. a finite length $A$-module) and the fact that the residue field of $A$ is a simple $\Lambda$-module implies $A$ is a finite-length $\Lambda$-module (and likewise for $B$). Why is this?

If we could show that the maximal ideal $m$ of $A$ is a finite-length $\Lambda$-module, then we'd be done. But how do you do this?

Answer 1

I think you meant to link this post, and that you're asking why $A$ is a finite length $\Lambda$-module.

As $A$ has finite length over itself, there exists a chain of ideals $(0)=I_0\subsetneq\cdots\subsetneq I_n=A$ such that for all $i\in\{1,\ldots,n\}$ we have that $I_i/I_{i-1}$ is a simple $A$-module. As $A$ is local, $\kappa(A)=A/\mathfrak{m}_A$ is up to isomorphism the only simple $A$-module. As $\kappa(A)$ is naturally isomorphic to $\kappa(\Lambda)=\Lambda/\mathfrak{m}_\Lambda$, the $A$-module sturcture is the same as the $\kappa(A)\cong\kappa(\Lambda)$-module structure on $I_i/I_{i-1}$, which in turn is the same as the $\Lambda$-module structure on $I_i/I_{i-1}$. Hence $(0)=I_0\subsetneq\cdots\subsetneq I_n=A$ is also a composition series for $A$ as a $\Lambda$-module, and thus $A$ is a $\Lambda$-module of finite length.