Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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".

share|cite|improve this question
up vote 3 down vote accepted

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 $\Lambda$.

share|cite|improve this answer
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. – Martin Brandenburg May 3 '11 at 14:19
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. – Jonathan Wise Jul 14 at 16:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.