Let $A,B,C$ be local artinian rings and $p : A \to C, q : B \to C$ local homomorphisms. Why is the fiber product $A \times_C B$ again a local artinian ring?

It is easy to see that $P:=A \times_C B$ is a local ring with maximal ideal $\mathfrak{m}_P=pr_A^{-1}(\mathfrak{m}_A) = pr_B^{-1}(\mathfrak{m}_B)$. Since some power of $\mathfrak{m}_A$ vanishes, the same is true for $\mathfrak{m}_P$. Thus it suffices to prove that $\mathfrak{m}_P$ is finitely generated. But I don't know how to find generators. Is this true at all?

Remark: This is used (without proof) in Schlessinger's article "Functors on Artin rings".


1 Answer 1


In the article the author considers $\Lambda$-algebras (where $\Lambda$ is a complete local noetherian ring) which are local, Artin and have the same residue field as $\Lambda$ (which means that the composition $\Lambda \rightarrow A \rightarrow A/\mathfrak{m}_A$ is surjective).

In that case $A$ is a $\Lambda$-module of finite length (because simple $A$-modules are $k$-lines hence simple $\Lambda$-modules, and $A$ is an $A$-module of finite length), and similarly for $B$, so $A \times_C B \subset A \times B$ is also a $\Lambda$-module of finite length, and so it is an artinian $\Lambda$-algebra.

Edit: for those who do not have access to the article, $k$ denotes the residue field of $\Lambda$.

  • $\begingroup$ Ah, thank you! So perhaps it's false in the general case? Anyway I accept this answer because this is the only case which is needed in the article. $\endgroup$ May 3, 2011 at 14:19
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    $\begingroup$ Some kind of residue field assumption is necessary. For example, $A = \mathbf C(x)[\epsilon] / (\epsilon^2)$, $B = \mathbf C(y)[\delta] / (\delta^2)$, $C = \mathbf C(x,y)$. The fiber product is $\mathbf C + \mathbf C(x) \epsilon + \mathbf C(y) \delta$, which is not artinian. $\endgroup$ Jul 14, 2016 at 16:06

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