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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.


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\ $?

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How hard could it be to capitalize the first letter in a sentence? Not starting a sentence with "about" is extra bonus for grammar. – Victor Protsak Jul 27 '10 at 6:51
No: $\mathbf{F}_ p[[u]] \otimes_ {\mathbf{Z}_p} \mathbf{F}_ p[[v]]$ is not local ($1 + u \otimes v$ is non-unit, since non-unit in $\mathbf{F}_ p[[u]] \otimes \mathbf{F}_ p((v))$ via computing under injection to $(\mathbf{F}_ p((v)))[[u]]$, so lies in extra max. ideal), nor noetherian. If $A$, $B$, and $C$ are complete loc. noeth. and $C \rightrightarrows A, B$ are local maps inducing isoms (or purely insep. finite ext'ns) on residue fields then completed tensor product $A \widehat{\otimes}_C B$ is complete loc. noeth. (and a categorical coproduct as such). See EGA 0$_{\rm{IV}}$, – BCnrd Jul 27 '10 at 10:00
About Victor's comment, let me just say that many titles of MO questions, including this one, are not meant to be sentences. – Tom Goodwillie Jul 27 '10 at 10:04
up vote 5 down vote accepted

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 $(1-tu)^{-1}=\sum t^ku^k$.)

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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 $\mathbb{F}_p$.

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If you ask that both $A$ and $B$ be integral over $\mathbb{Z}_p$, then it's true that $A\otimes B$ will be local. You can get this pretty easily using the "going up/going down" theorems. – Joel Dodge Jul 27 '10 at 5:49
The residue field of $B=\mathbb{Q}_p$ is not $F_p$. I want $A$ and $B$ with the residue field $F_p$, the finite field with $p$ elements. – N. Kumar Jul 27 '10 at 5:52
ah right of course i forgot that while thinking about this. . . – Joel Dodge Jul 27 '10 at 5:53

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