$\mathfrak a$-adic Completion of a Ring is Flat

Let $$A$$ be a Noetherian ring and $$\mathfrak a \subset A$$ an ideal of $$A$$. Denote by $$\hat{A_{\mathfrak a}}:= \varprojlim_n A/\mathfrak a^n$$ the $$\mathfrak a$$-adic completion of $$A$$ wrt $$\mathfrak a$$.

Why and how to see that $$\hat{A_{\mathfrak a}}$$ is a flat $$A$$-module?

Could anybody give sketch for the proof?

My considerations: up to now I know that firstly $$\mathfrak a$$ is finitely generated.

Remark: As @Max mentioned below one cannot expect that the canonical map $$A \to \hat{A_{\mathfrak a}}$$ is injective (Krull's intersection thm only works for local Noetherians).

• Krull's intersection theorem is for local rings; there are examples where $A\to \widehat{A}_a$ is not injective, even for noetherian rings. – Max Feb 17 at 20:01
• @Max: thank you for this important remark. – KarlPeter Feb 18 at 11:59

• You can find a solution to this problem in the following document of Conrad Artin–Rees and completions. In particular it is the Corollary 3.2.

• Also, in the book of Qing Liu - Algebraic Geometry and Arithmetic Curves you can find a proof in the Theorem 3.15. I love how Liu wrote the review of commutative algebra in his book and I recommend you to read it.

Here a snap of the proof taken from Liu's book.

The Corollary 3.14 that Liu is referring to is the well-known result:

• Here the approach to this theorem in Matsumura: Commutative Ring Theory, Chapter 3.

The main theorem is Theorem 8.8:

The theorems that Matsumura is referring to are:

Theorem 7.7:

Theorem 8.1:

Theorem 8.6:

Note The notation of the above theorem is: $$A$$ a noetherian ring, $$M$$ a finite $$A$$-module, $$N \subset M$$ a submodule, and $$I$$ an ideal of $$A$$.

I hope this helps you. If you don't understand the references I gave you, don't hesitate to ask!

• I have two questions: 1) the "canonically" morphism $M \otimes_A \hat{A} \to \hat{M}$ from 3.14 is given concretely via $m \otimes (\alpha_n)_n \mapsto (m \cdot\alpha_n)_n$? As well as $\hat{J} \to \hat{A}$ componentswise $J/JI^n \to A/I^n, \iota + JI^n \mapsto + \iota +I^n$ in each $n$? – KarlPeter Feb 20 at 2:19
• 2) Another point regarding proof of thm 3.15 is why is $\beta_m \in I^nJ$? – KarlPeter Feb 20 at 2:19
• 1) The mapping $M\otimes_A \hat{A} \rightarrow \hat{M}$ is by definition called canonical (Bourbaki, Commutative Algebra, pg. 202) It's is defined as you said, by the universal property of tensor products applied to the map $M \times \hat{A} \rightarrow \hat{M} \ : \ (m,(a_n)_n) \mapsto (a_n \cdot m)_n.$ Yes, the map $\hat{J} \rightarrow \hat{A}$ is defined this way. Note that it's the map induced by the $A$-linear map $J \rightarrowtail A,$ the injection of $J$ in $A.$ – DrinkingDonuts Feb 20 at 10:03
• 2) If $\beta_m + I^{m}J = \alpha_m + I^{m}J$ then $\beta_m - \alpha_m \in I^{m}J$ and $I^{m}J \subset I^{m} \cap J \subset I^{m}.$ Furthermore, since $\alpha$ is in the kernel, we get that $\alpha_m \in I^{m}$ so $\beta_m \in I^{m}$ and $\beta_m \in I^{m} \cap J \subset I^{n}J.$ – DrinkingDonuts Feb 20 at 10:17
• ah ok and $\alpha_n$ MUST be image of $\beta_m$ since all elements $(\alpha_n)_n \in \hat{J}$ are compatible in the sense that $\alpha_m + JI^m \mapsto \alpha_n + JI^n$ for $m \ge n$ by definition. So the image of $\beta_m$ in $J/JI^n$ doesn't have "another choice" as to be $\alpha_n$? – KarlPeter Feb 20 at 17:06

The Stacks project has a proof, https://stacks.math.columbia.edu/tag/06LE, considering the direct sum over a one element index set.