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This is a follow-up to About the tensor product of two local rings.
Let $A, B$ and $C$ be local rings (commutative and noetherian). Suppose that we have local ring maps $C \to A$ and $C \to B$.
What assumptions guarantee that the tensor product $A \otimes_C B$ of $A$ and $B$ over $C$ is a local ring?

For example, $A$ is the completion of $C$? What if $A, B$ and $C$ are dvrs? Are there general criteria? (If I remember correctly, I have seen one criterion, when $C$ was a field. What about general $C$?)

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Are you sure this holds when C is a field? –  J.C. Ottem Feb 2 '11 at 18:28
Not, of course (many trivial counter-examples). I said that there is known (if I remember correctly) criterion for when such tensor is local, if we assume that C is a field. –  unknown Feb 2 '11 at 18:37
Have you seen jstor.org/stable/2040682?seq=1? –  J.C. Ottem Feb 2 '11 at 19:01
Yes. But it considers the case, when C is a field. –  unknown Feb 2 '11 at 19:03
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1 Answer

Here is a partial answer related to connected rings (which is more general than local rings) in the case where rings are not necessarily commutative.

If $K$ is a field, let $A$ be an artinian and connected $K$-algebra (in this situation it's equivalent to being local and artinian). Assume that $A$ is endowed with a morphism of algebras $A\longrightarrow K$. Then for any connected $K$-algebra $B$, $A\otimes_K B$ is connected.

An trivial counterexample : if $K$ is $\mathbb{F}_2$ and $L=\mathbb{F}_4$, $L\otimes_K L$ is not connected (sorry to write it this way but there seems to be a bug with latex).

edit : this results is also true if $K$ is only supposed to be a domain and with some assumption on $B$ (flatness).

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