Product of reduced affinoid spaces over a field is reduced (reference request)

Question

Let $K$ be a field of characteristic zero complete with respect to a non-Archimedean absolute value. Suppose that $A$ and $B$ are two affinoid $K$-algebras. I'd like a reference that will answer the following question:

If $A$ and $B$ are both reduced then is the completed tensor product $A\widehat{\otimes}_K B$ also reduced?

In a more algebraic situation, the answer is well-known (e.g. it is discussed here and here) to have an affirmative answer: if $A$ and $B$ are reduced $K$-algebras then the tensor product $A\otimes_K B$ is also reduced. But, the proof does not seem to directly carry over.

Geometrically I'm just asking if the fibered product of two reduced $K$-affinoid spaces is reduced, so surely someone has thought about this before. I tried searching through the famous book of Bosch, Güntzer and Remmert but didn't find anything. The papers of Bosch and Lütkebohmert (and Raynaud) had no similar fact either.

Answer 1

For all those kind of properties, I recommend looking at Ducros's paper "Les espaces de Berkovich sont excellents" (Ann. Inst. Fourier 59 (2009), no. 4, 1407-1516, http://aif.cedram.org/item?id=AIF_2009__59_4_1443_0). It is written in the language of Berkovich spaces, so it is even slightly more general than what you want.

More precisely, théorème 8.1 tells you that the product of two geometrically reduced spaces is geometrically reduced. Note also that in your case (and more generally over a perfect field), reduced and geometrically reduced are equivalent (see proposition 6.3 and remarque 6.5).

Finally, to answer precisely your question, you also need to know that an affinoid algebra $A$ is reduced if, and only if, the associated affinoid space $\mathcal{M}(A)$ (or whatever notation you like best depending on which theory you use) is. In one direction, this is clear: if the space is reduced, then so is its ring of global functions.

The other direction is more difficult and relies on the excellence of the affinoid algebra $A$ (thanks to nfdc23 for pointing out that I had forgotten this condition). Using this, in the strictly analytic case (which is the one you want to consider in the setting of rigid geometry), you can deduce that the completions of the analytic local rings at rigid points are reduced because those completions coincide with that of the algebraic local rings (and that the algebraic local rings are reduced and excellent). From this, you deduce that the analytic local rings at rigid points are reduced, hence that the algebras of the affinoid domains are reduced, hence that every analytic local ring is reduced, since it is a colimit of the latter (see theorem 2.2.1 in Berkovich's IHES paper).

Answer 2

Jerome Poineau is correct that Ducros' work on "homological properties" of Berkovich spaces gives the most robust and comprehensive approach to these matters. I will give a more direct geometric argument in the setting of rigid geometry to avoid digging far into his paper (wonderful as it is!).

The question posed is a special case of behavior for geometric reducedness (appropriately defined!) in the setting of affinoid algebras, equivalent to reducedness when the ground field is perfect (e.g., characteristic 0) by Lemma 3.3.1 in the paper "Irreducible Components of Rigid Spaces". So it is more natural to pose the question for geometric reducedness and aim to show that geometric reducedness is preserved under such completed tensor products. We may and do extend scalars to a completed algebraic closure of $K$ so that $K$ is algebraically closed. Hence, now regularity and smoothness at a point of the rigid space are equivalent (as each may be checked on the completed local ring, which coincides with the completion of the corresponding algebraic local ring).

By excellence of affinoid algebra, the regular locus is Zariski-open and dense in the spectrum of any reduced $K$-affinoid algebra, so there are elements $a \in A$ and $b \in B$ that are not zero-divisors such that the rings $A[1/a]$ and $B[1/b]$ are regular. In particular, by considering completed local rings on the analytic side, we see that $U = \{a \ne 0\}$ and $V = \{b \ne 0\}$ in $X := {\rm{Sp}}(A)$ and $Y := {\rm{Sp}}(B)$ respectively are regular, or equivalently smooth. Hence, $U \times V$ is smooth, so any nilpotent in the ring $A \widehat{\otimes}_K B$ of global functions on $X \times Y$ vanishes over $U \times V$.

Thus, it suffices to show that restriction to $U \times V$ is injective on the ring of global function on $X \times Y$. If we were in usual algebraic geometry this would be easy since only usual tensor products would intervene, but here we have to contend with completed tensor products. Observe that $U \times V$ is the complement of the vanishing locus of $f = a \widehat{\otimes} b \in A \widehat{\otimes}_K B$, and $f$-multiplication is injective on $A \widehat{\otimes}_K B$ (since in general completed tensor product of countable type $K$-Banach algebras carries continuous injections with closed image to injections; shouldn't need countable type, I think due to Lazard, but whatever).

Now our problem is reduced (ha-ha) to showing that if $C$ is $K$-affinoid and $f \in C$ is not a zero-divisor then for $Z := {\rm{Sp}}(C)$, $W := \{f \ne 0\} \subset Z$, if $c \in C$ vanishes on $W$ then $c = 0$. Better yet, without any hypotheses on $f$ not being a zero-divisor, if $c|_W = 0$ then we claim that $f^n c = 0$ in $C$ for some $n \ge 0$. Consider the coherent annihilator ideal sheaf $\mathscr{I} = {\rm{Ann}}_{O_Z}(c)$; this is associated to the ideal $I = {\rm{Ann}}(c) \subset C$. By hypotheses $\mathscr{I}|_W = O_W$, so ${\rm{Sp}}(C/I)$ does not meet $W$, which is to say it is contained in $Z-W = {\rm{Sp}}(C/(f))$ as analytic sets inside $Z$. Hence, by the Jacobson property of affinoid algebras (aka "analytic Nullstellensatz"), some power of $(f)$ is contained in $I$, which is to say $f^n \in I$ for some $n \ge 0$; i.e., $f^n c = 0$ in $C$ as desired.