@SamHopkins- Thanks! No I don't have a counterexample for infinite posets, but I'm very interested in the infinite case as well. I thought it was natural to concentrate first on the finite case.

@TimCampion - Not sure what it really means, but I did this Ngram search books.google.com/ngrams/…

Rahman. M replaced the tag ultrapowers with set-theory. I rejected the edit because the question asks if certain pairs of ultrapowers are isomorphic. If I had to keep just one tag, I would choose this one. It seems to me the other tags I picked give a more precise idea of the question than the set-theory tag. If you think the ultrapowers tag is inappropriate, thanks for telling me why.

Crossposted: math.stackexchange.com/q/3411075/660

I realize that I shouldn't have used the notation $p(x)$ for an element of $R$, since it suggests that it's a polynomial in $x$, instead of a polynomial in $x,1/(x-1),1/(x-2),...$

I'm the one who's sorry! I'm most grateful for your awesome answer and your comments! I convinced myself that $R\to R\times k$ is an epi by checking the identity $(p(x),\lambda)\otimes(1,1)=(1,1)\otimes(p(x),\lambda)$ in $(R\times k)\otimes_R(R\times k)$. I hope this argument is correct. Even if it is, I'm sure your approach is more elegant, but I'm not familiar with schemes (to say the least!). I would guess that scheme theory helped you find your example more that it was needed for your proof, but I may be completely wrong...

Thanks! I'm confused. You write: "$f:A\to B$ is an epimorphism if and only if $B\otimes_AB\to B$ is an isomorphism; in particular this is also equivalent to $\operatorname{Spec}B\to\operatorname{Spec} A$ being a monomorphism". I don't understand the second phrase: if $A$ and $B$ are fields and $f$ is not surjective, then $f$ is not an epimorphism but $\operatorname{Spec}B\to\operatorname{Spec} A$ is a monomorphism, right?