# Is $A[[x_1,x_2,\dots]]$ flat over $A[x_1,x_2,\dots]$? ($A$ noetherian and commutative)

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.

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.

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.

• 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? – user26857 Apr 2 at 15:06
• @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. – Pierre-Yves Gaillard Apr 2 at 15:44