Descent of flatness from algebras to monoids Consider a morphism of commutative monoids $u\colon M\rightarrow N$. We say that $u$ is flat, if the tensor product functor $\bullet\otimes_MN$ from the category of $M$-modules to the category of $N$-modules commutes with finite projective limits.
Let $R$ be a non-zero commutative ring, and suppose that the induced morphism of $R$-algebras $R[u]\colon R[M]\rightarrow R[N]$ is flat.

Can we conclude that $u\colon M\rightarrow N$ is flat?

 A: No.  Take $R$ to be the field of rationals $\mathbb Q$. Let $M$ be the multiplicative monoid of the two element field and $N$ the additive group of this field. Let $u$ be the trivial homomorphism. Both monoid algebras are isomorphic to $\mathbb Q\times \mathbb Q$ so semisimple. Thus all modules over both rings are projective hence flat  But $u$ is not flat since it doesn't preserve the terminal object. 
More conceptually, for $u$ to be flat means that the category of elements of $N$ as an $M$-set is filtering.  By Quillen's Theorem A this would make the classifying spaces of $M$ and $N$ homotopy equivalent.  But the classifying space of $M$ is contractible and the classifying space of $N$ is a $K(\pi,1)$ for the two element group.
A: Here is an easier answer.  Let $R$ be any non-zero commutative ring with unit.  Let $G$ be any group and let $H$ be a proper subgroup.  Let $u\colon H\to G$ be the inclusion.  Then $u$ is never flat but $RG$ is a free $RH$-module and hence flat.  The problem is that the terminal $H$-set $1$ tensored with $G$ over $H$ yields $G/H$ as a $G$-set, which is not there terminal $G$-set. 
