why "$B[[u]]$ is flat over $B \otimes_A A[[u]]$ "?

Question

Lemma:

If $A$ is a Noetherian and $A \to B$ is an fppf morphism, then the induced $A[[u]] \to B[[u]]$ is faithfully flat.

Proof:

Note that $B[[u]]$ is flat over $B \otimes_A A[[u]]$ (being $u$-adic completion of the latter ring, which is finitely presented over the Noetherian ring $A[[u]]$, and thus is Noetherian itself), which in turn flat over $A[[u]]$. ( proof continues$\cdots \cdots$)

This is collected from a paper.

But I did not understand why "$B[[u]]$ is flat over $B \otimes_A A[[u]]$ " ?

Because here $B[[u]]$ is a ring and so how to describe its flatness?

I am thinking, that if $M,N$ be two $B \otimes_A A[[u]]$-module and if $P$ be any $B[[u]]$-module and if $\phi:M \to N$ be injective then $M \otimes_{B \otimes_A A[[u]]} P \to N \otimes_{B \otimes_A A[[u]]} P$ is also injective. That means I am trying to show that every module over $B[[u]]$ is flat and hence it is flat ring. Is it true what I am saying?

How to see that $B[[u]]$ is an $u$-adic completion of $B \otimes_A A[[u]]$ ?

Kindly help me an explanation why "$B[[u]]$ is flat over $B \otimes_A A[[u]]$ " ?

Answer 1

If you're unaware of how one ring may be flat over another ("[b]ecause here $B[[u]]$ is a ring and so how to describe its flatness?"), it's probably a good idea to review that first:

A ring homomorphism $f:R\to S$ gives $S$ the structure of an $R$-module by defining the $R$-action on $S$ as $r\cdot s = f(r)s$ where the multiplication on the right is the multiplication in $S$. We then declare $f$ to make $S$ a flat $R$-module if $S$ is flat as an $R$-module via the module structure given by $f$.

The correct strategy to resolve your question is to recall a foundational result from commutative algebra:

Lemma(Stacks 00MB): Let $R$ be a ring and $I$ be an ideal. Assume $R$ is noetherian. Then $R^\wedge$, the completion of $R$ with respect to $I$, is a flat $R$-module.

This means that if we can show that $B[[u]]$ is (isomorphic to) the completion of $B\otimes_A A[[u]]$ with respect to the ideal $(1\otimes u)$, then we'll be done. This is straightforwards: the completion $R^\wedge$ is defined as the inverse limit of $R/I^n$ as $n$ varies. It is straightforwards to see that $(B\otimes_A A[[u]])/(1\otimes u)^n\cong B\otimes_A (A[u]/(u^n))\cong B[u]/(u^n)$ which clearly has inverse limit $B[[u]]$. Thus $B[[u]]$ is the completion of $B\otimes_A A[[u]]$ and is flat over it.