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Let $A$ be a noetherian commutative ring with one and $x_1,x_2,\dots$ indeterminates.

Question. Is $A[[x_1,x_2,\dots]]$ flat over $A[x_1,x_2,\dots]\ ?$

Recall that $A[[x_1,x_2,\dots]]$ is the set of all formal expressions $\sum a_uu$ where $u$ runs over the set of monomials in $x_1,x_2,\dots$ and $a_u\in A$, the ring structure being the obvious one.

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Here is a new version of the answer. The comments refer to the previous version, which is pasted below.

The answer is Yes, that is

(a) $A[[x_1,x_2,\dots]]$ is flat over $A[x_1,x_2,\dots]$.

Claim 1: $A[[x_1,x_2,\dots]]$ is flat over $A[x_1,\dots,x_n]$.

Claim 2: Claim 1 implies (a).

Proof of Claim 2. Set $$ B_n:=A[[x_1,x_2,\dots]] \otimes_{A[x_1,\dots,x_n]}A[x_1,x_2,\dots]. $$ Then $B_n$ is flat over $A[x_1,x_2,\dots]$. Moreover $A[[x_1,x_2,\dots]]$ is the colimit of the $B_n$. As filtered colimits preserve flatness, Claim 2 is proved.

Proof of Claim 1. The ring $A[[x_1,\dots,x_n]]$ being noetherian by Lemma 10.30.2 of [1], and flat over $A[x_1,\dots,x_n]$ by Lemma 10.96.2(1) of [1], it is enough to verify that $A[[x_1,x_2,\dots]]$ is flat over $A[[x_1,\dots,x_n]]$.

But, since $A[[x_1,x_2,\dots]]$, viewed as an $A[[x_1,\dots,x_n]]$-module, is just a product of copies of $A[[x_1,\dots,x_n]]$, it is flat over $A[[x_1,\dots,x_n]]$ by Lemma 10.89.5 and Proposition 10.89.6 of [1], we are done.

[1] The Stacks Project.

An obvious variant of the above argument shows:

If $S$ is an arbitrary set of indeterminates, then $A[[(x)_{x\in S}]]$ is flat over $A[(x)_{x\in S}]$.


Previous version:

The answer is Yes, that is

(a) $A[[x_1,x_2,\dots]]$ is flat over $A[x_1,x_2,\dots]$.

Claim 1: $A[[x_1,x_2,\dots]]$ is flat over $A[x_1,\dots,x_n]$.

Claim 2: Claim 1 implies (a).

Proof of Claim 2. Set $$ B_n:=A[[x_1,x_2,\dots]] \otimes_{A[x_1,\dots,x_n]}A[x_1,x_2,\dots]. $$ Then $B_n$ is flat over $A[x_1,x_2,\dots]$. Moreover $A[[x_1,x_2,\dots]]$ is the colimit of the $B_n$. As filtered colimits preserve flatness, Claim 2 is proved.

Proof of Claim 1. The ring $A[[x_1,\dots,x_n]]$ being noetherian by Lemma 10.30.2 of [1], and flat over $A[x_1,\dots,x_n]$ by Lemma 10.96.2(1) of [1], it is enough to verify that $A[[x_1,x_2,\dots]]$ is flat over $A[[x_1,\dots,x_n]]$.

But, since $A[[x_1,x_2,\dots]]$, viewed as an $A[[x_1,\dots,x_n]]$-module, is isomorphic to $$ (A[[x_1,\dots,x_n]])[[x_{n+1}]], $$ which is flat over $A[[x_1,\dots,x_n]]$ by Lemma 10.89.5 and Proposition 10.89.6 of [1], we are done.

[1] The Stacks Project.

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  • $\begingroup$ I don't understand this part: "since $A[[x_1,x_2,\dots]]$, viewed as an $A[[x_1,\dots,x_n]]$-module, is isomorphic to $(A[[x_1,\dots,x_n]])[[x_{n+1}]]$..." Where are the variables from $x_{n+2}$ on? $\endgroup$ – user26857 Apr 2 at 15:06
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    $\begingroup$ @user26857 - As a $B$-module, $B[[t]]$ is just the product of countably many copies of $B$, and similarly for $B[[x_1,\dots,x_n]]$ and $B[[x_1,x_2,\dots]]$. In all cases, there are only countably many monomials. Right? - Also $A[[x_1,x_2,\dots]]$ is isomorphic to $$(A[[x_1,\dots,x_n]])[[x_{n+1},x_{n+2},\dots]]$$ (even as a ring), but I think we agree on this. $\endgroup$ – Pierre-Yves Gaillard Apr 2 at 15:44

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