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:



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> If $A$ and $B$ are both reduced then is the completed tensor product $A\widehat{\otimes}\_K B$ also reduced?
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In a more algebraic situation, the answer is well-known (e.g. it is discussed [here](https://mathoverflow.net/questions/9436/reduced-%E2%8A%97-reduced-reduced-what-about-connected "here") and [here](https://math.stackexchange.com/questions/308015/how-to-show-the-tensor-product-of-reduced-k-algebras-must-be-reduced "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.



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


