Is tensor product of local algebras local?

Question

In general, the tensor product of two local rings is not local. For example, $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}\ $ is not a local ring.

Let $\mathbb{F}_{p}$ denote the finite field with $p$ elements. Let $A,B$ be two complete local noetherian $\mathbb{Z}_p$-algebras with residue field $\mathbb{F}_p$. Let $m_A, m_B$ denote the maximal ideals of $A,B$, respectively.

Question:

Is it true that $A \otimes_{\mathbb{Z}_p} B\ $ is a local ring?

Clearly, the ideal $m_A \otimes B + A \otimes m_B$ is a maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $ with the residue field $F_p$. Is it the only maximal ideal of $A \otimes_{\mathbb{Z}_p} B\ $?

Answer 1

Let $A=\mathbb F_p[[t]],B=\mathbb F_p[[u]]$. Then, $1\otimes1-t\otimes u$ is neither in $\mathfrak m_A\otimes B+A\otimes\mathfrak m_B$ nor a unit, so it is contained in some other maximal ideal of $A\otimes B$. (Proof that $1\otimes1-t\otimes u$ is not a unit: An element of $\mathbb F_p[[t]][[u]]$ coming from $A\otimes B$ has the property that its coefficients with respect to $u$ span a finite-dimensional subspace of $\mathbb F_p[[t]]$, but this fails for the coefficients $t^k$ of $(1-tu)^{-1}=\sum t^ku^k$.)

Answer 2

Let $P$ be a finite $p$-group, let $A = \mathbb{Z}_p[P]$ be the group ring and let $B=\mathbb{Q}_p$. Then $A$ and $B$ are both complete Noetherian local $\mathbb{Z}_p$-algebras but $A\otimes B=\mathbb{Q}_p[P]$ is a product of fields, hence not local.

Edit: This fails to be a counterexample because the residue field of $\mathbb{Q}_p$ is not $\mathbb{F}_p$.