Lemma 110.12.1. Let $R$ be a ring. Let $M$ be an $R$-module which is countable. Then $M$ is a finite $R$-module if and only if $M \otimes _ R R^\mathbf {N} \to M^\mathbf {N}$ is surjective.

## 110.12 Nonflat completions

The completion of a ring with respect to an ideal isn't always flat, contrary to the Noetherian case. We have seen two examples of this phenomenon in More on Algebra, Example 15.90.9. In this section we give two more examples.

**Proof.**
If $M$ is a finite module, then the map is surjective by Algebra, Proposition 10.89.2. Conversely, assume the map is surjective. Let $m_1, m_2, m_3, \ldots $ be an enumeration of the elements of $M$. Let $\sum _{j = 1, \ldots , m} x_ j \otimes a_ j$ be an element of the tensor product mapping to the element $(m_ n) \in M^\mathbf {N}$. Then we see that $x_1, \ldots , x_ m$ generate $M$ over $R$ as in the proof of Algebra, Proposition 10.89.2.
$\square$

Lemma 110.12.2. Let $R$ be a countable ring. Let $M$ be a countable $R$-module. Then $M$ is finitely presented if and only if the canonical map $M \otimes _ R R^\mathbf {N} \to M^\mathbf {N}$ is an isomorphism.

**Proof.**
If $M$ is a finitely presented module, then the map is an isomorphism by Algebra, Proposition 10.89.3. Conversely, assume the map is an isomorphism. By Lemma 110.12.1 the module $M$ is finite. Choose a surjection $R^{\oplus m} \to M$ with kernel $K$. Then $K$ is countable as a submodule of $R^{\oplus m}$. Arguing as in the proof of Algebra, Proposition 10.89.3 we see that $K \otimes _ R R^\mathbf {N} \to K^\mathbf {N}$ is surjective. Hence we conclude that $K$ is a finite $R$-module by Lemma 110.12.1. Thus $M$ is finitely presented.
$\square$

Lemma 110.12.3. Let $R$ be a countable ring. Then $R$ is coherent if and only if $R^\mathbf {N}$ is a flat $R$-module.

**Proof.**
If $R$ is coherent, then $R^\mathbf {N}$ is a flat module by Algebra, Proposition 10.90.6. Assume $R^\mathbf {N}$ is flat. Let $I \subset R$ be a finitely generated ideal. To prove the lemma we show that $I$ is finitely presented as an $R$-module. Namely, the map $I \otimes _ R R^\mathbf {N} \to R^\mathbf {N}$ is injective as $R^\mathbf {N}$ is flat and its image is $I^\mathbf {N}$ by Lemma 110.12.1. Thus we conclude by Lemma 110.12.2.
$\square$

Let $R$ be a countable ring. Observe that $R[[x]]$ is isomorphic to $R^\mathbf {N}$ as an $R$-module. By Lemma 110.12.3 we see that $R \to R[[x]]$ is flat if and only if $R$ is coherent. There are plenty of noncoherent countable rings, for example

where $k$ is a countable field. This ring is not coherent because the ideal $(y, z)$ of $R$ is not a finitely presented $R$-module. Note that $R[[x]]$ is the completion of $R[x]$ by the principal ideal $(x)$.

Lemma 110.12.4. There exists a ring such that the completion $R[[x]]$ of $R[x]$ at $(x)$ is not flat over $R$ and a fortiori not flat over $R[x]$.

**Proof.**
See discussion above.
$\square$

It turns out there is a ring $R$ such that $R[[x]]$ is flat over $R$, but $R[[x]]$ is not flat over $R[x]$. See this post by Badam Baplan. Namely, let $R$ be a valuation ring. Then $R$ is coherent (Algebra, Example 10.90.2) and hence $R[[x]]$ is flat over $R$ by Algebra, Proposition 10.90.6. On the other hand, we have the following lemma.

Lemma 110.12.5. Let $R$ be a domain with fraction field $K$. If $R[[x]]$ is flat over $R[x]$, then $R$ is normal if and only if $R$ is completely normal (Algebra, Definition 10.37.3).

**Proof.**
Suppose we have $\alpha \in K$ and a nonzero $r \in R$ such that $r \alpha ^ n \in R$ for all $n \geq 1$. Then we consider $f = \sum r \alpha ^{n - 1} x^ n$ in $R[[x]]$. Write $\alpha = a/b$ for $a, b \in R$ with $b$ nonzero. Then we see that $(a x - b)f = -rb$. It follows that $rb$ is in the ideal $(ax - b)R[[x]]$. Let $S = \{ h \in R[x] : h(0) = 1\} $. This is a multiplicative subset and flatness of $R[x] \to R[[x]]$ implies that $S^{-1}R[x] \to R[[x]]$ is faithfully flat (details omitted; hint: use Algebra, Lemma 10.39.16). Hence

is injective. We conclude that $h rb = (ax - b) g$ for some $h \in S$ and $g \in R[x]$. Writing $h = 1 + h_1 x + \ldots + h_ d x^ d$ shows that we obtain

This factorization in $K[x]$ gives a corresponding factorization in $K[x^{-1}]$ which shows that $\alpha $ is the root of a monic polynomial with coefficients in $R$ as desired. $\square$

Lemma 110.12.6. If $R$ is a valuation ring of dimension $> 1$, then $R[[x]]$ is flat over $R$ but not flat over $R[x]$.

**Proof.**
The arguments above show that this is true if we can show that $R$ is not completely normal (valuation rings are normal, see Algebra, Lemma 10.50.3). Let $\mathfrak p \subset \mathfrak m \subset R$ be a chain of primes. Pick nonzero $x \in \mathfrak p$ and $y \in \mathfrak m \setminus \mathfrak p$. Then $x y^{-n} \in R$ for all $n \geq 1$ (if not then $y^ n/x \in R$ which is absurd because $y \not\in \mathfrak p$). Hence $1/y$ is almost integral over $R$ but not in $R$.
$\square$

Next, we will construct an example where the completion of a localization is nonflat. To do this consider the ring

Denote $f \in R$ the residue class of $z$. We claim the ring map

isn't flat. Let $I$ be the kernel of $y : R[[x]] \to R[[x]]$. A typical element $g$ of $I$ looks like $g = \sum g_{n, m} a_ mx^ n$ where $g_{n, m} \in k[z]$ and for a given $n$ only a finite number of nonzero $g_{n, m}$. Let $J$ be the kernel of $y : R_ f[[x]] \to R_ f[[x]]$. We claim that $J \not= I R_ f[[x]]$. Namely, if this were true then we would have

for some $m \geq 1$, $g_ i \in I$, and $h_ i \in R_ f[[x]]$. Say $h_ i = \bar h_ i \bmod (y, a_1, a_2, a_3, \ldots )$ with $\bar h_ i \in k[z, 1/z][[x]]$. Looking at the coefficient of $a_ n$ and using the description of the elements $g_ i$ above we would get

for some $\bar g_{i, n} \in k[z][[x]]$. This would mean that all $z^{-n}x^ n$ are contained in the finite $k[z][[x]]$-module generated by the elements $\bar h_ i$. Since $k[z][[x]]$ is Noetherian this implies that the $R[z][[x]]$-submodule of $k[z, 1/z][[x]]$ generated by $1, z^{-1}x, z^{-2}x^2, \ldots $ is finite. By Algebra, Lemma 10.36.2 we would conclude that $z^{-1}x$ is integral over $k[z][[x]]$ which is absurd. On the other hand, if (110.12.6.1) were flat, then we would get $J = IR_ f[[x]]$ by tensoring the exact sequence $0 \to I \to R[[x]] \xrightarrow {y} R[[x]]$ with $R_ f[[x]]$.

Lemma 110.12.7. There exists a ring $A$ complete with respect to a principal ideal $I$ and an element $f \in A$ such that the $I$-adic completion $A_ f^\wedge $ of $A_ f$ is not flat over $A$.

**Proof.**
Set $A = R[[x]]$ and $I = (x)$ and observe that $R_ f[[x]]$ is the completion of $R[[x]]_ f$.
$\square$

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