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).

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

Theorem 3.15 - From Liu's book.

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

Corollary 3.14 - From Liu's book.

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

The main theorem is Theorem 8.8:

Theorem 8.8 - From Matsumura's book.

The theorems that Matsumura is referring to are:

Theorem 7.7:

Theorem 7.7 - From Matsumura's book.

Theorem 8.1:

Theorem 8.1 - From Matsumura's book.

Theorem 8.6:

Theorem 8.6 - From Matsumura's book.

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!

  • $\begingroup$ 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 $n$? $\endgroup$ – KarlPeter Feb 20 at 2:19
  • $\begingroup$ 2) Another point regarding proof of thm 3.15 is why is $\beta_m \in I^nJ$? $\endgroup$ – KarlPeter Feb 20 at 2:19
  • 1
    $\begingroup$ 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 $A.$ $\endgroup$ – DrinkingDonuts Feb 20 at 10:03
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    $\begingroup$ 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 $\beta_m \in I^{m} \cap J \subset I^{n}J.$ $\endgroup$ – DrinkingDonuts Feb 20 at 10:17
  • $\begingroup$ 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 $\alpha_n$? $\endgroup$ – 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.


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