Is the completed tensor product (over a complete dvr) of two reduced complete Noetherian local rings again reduced?

Question

To be more specific, Let $\mathcal{O}$ be a finite extension of $\mathbb{Z}_{p}$. Let $A=\mathcal{O}[[X_{1},\ldots, X_{n}]]/\left( f_{1},\ldots,f_{r}\right) $ and $B=\mathcal{O}[[Y_{1},\ldots, Y_{m}]]/\left( g_{1},\ldots,g_{s}\right)$ be quotients of power series rings in several variables over $\mathcal{O}$ by closed ideals generated by certain power series $f_{i}$ and $g_{j}$. We assume $A$ and $B$ are both reduced (we can even assume one of them is a domain). Is it true that their completed tensor product $$ A \hat{\otimes}_{\mathcal{O}}B =\mathcal{O}[[X_{1},\ldots, X_{n},Y_{1},\ldots, Y_{m}]]/\left( f_{1},\ldots,f_{r},g_{1},\ldots,g_{s}\right) $$ is again reduced? (Edit: we do assume $A$ and $B$ are both $\mathcal{O}$-flat, for its applications in Galois deformation theory; see below.)

When one of the rings $A$ or $B$ is just a power series ring, this problem is not hard, as we can identify $ A \hat{\otimes}_{\mathcal{O}}B$ as a power series ring over a reduced ring, but it is considerably more involved when both of them are not. My current strategy is to prove that in this specific set-up, the sum of two radical ideals ($\left( f_{1},\ldots,f_{r}\right)$ and $\left( g_{1},\ldots,g_{s}\right)$) is again radical. But this does not seem to be straightforward. I also came across some general theorems about the reducedness of complete local rings, such as Theorem 7.9 of David Eisenbud's "Commutative algebra with a view toward algebraic geometry" (page 193, Section 7.4), but they don't seem to apply here either. There are other posts on MO about the reducedness of tensor products of reduced rings, but they are either about purely algebraic tensor products, or completed tensor products over (perfect) fields, which to the best of my knowledge cannot be extended to the above case in an obvious way.

This question arises in the context of Galois deformation theory (of $\operatorname{GL}_{2}\left( \mathbb{Q}\right)$, say), when we consider the tensor product of two universal deformations (so $A$ and $B$ are universal deformation rings). $A$ or $B$ being a power series ring means the corresponding deformation problem is unobstructed, which was proved by Tom Weston to be almost always the case. In general, it is a quotient of a power series ring, so we have to consider the situation described above.

Answer 1

Let $K$ be the fraction field of $\mathcal{O}$.

Assuming $A$ and $B$ are $\mathcal{O}$-flat, then also $A \widehat{\otimes} B$ is $\mathcal{O}$-flat (exercise), so it's enough to see that $C:=(A \widehat{\otimes} B)[1/p]$ is reduced. For this in turn it's enough to see that the completed local ring of $C$ at any maximal ideal is reduced. But now the rigid space $X=\mathrm{Spf}(A)_\eta \times_{\mathrm{Sp}K} \mathrm{Spf}(B)_\eta$ is reduced by one of the questions you referenced, and the completed local rings of $C$ at maximal ideals identify with the the completed local rings of $\mathcal{O}_X$ at closed points, which are reduced by the nice properties of rigid spaces mentioned in the same question you referenced.

There's probably an easier argument which avoids rigid geometry, but this is the cheapest thing I could come up with on my bus ride home. ;)