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Assume we are given Noetherian local rings $(A,\mathfrak{m})$ and $(B,\mathfrak{n})$ such that:

  • $A \subset B$ and $\mathfrak{m} = A \cap \mathfrak{n}$,
  • $B$ is a finite $A$-module.

It is known that the $\mathfrak{m}$-adic completion $\hat{A}$ is flat over $A$ and that the $\mathfrak{n}$-adic completion $\hat{B}$ is flat over $B$.

But, as $A$ is a subring of $B$, one may also look at the closure of $A$ in $\hat{B}$. The closure $\bar{A}$ is equal to the completion of $A$ with respect to the filtration $A \cap \mathfrak{n}^i$.

Are there any known results on when $\bar{A}$ is a flat over $A$?

Edit: There are examples when $\bar{A} = \hat{A}$; which would imply flatness of $\bar{A}$ over $A$. This happens when the induced topologies of the filtrations $\mathfrak{m}^i$ and $A \cap \mathfrak{n}^i$ are equal. Obviously $\mathfrak{m}^i \subset A \cap \mathfrak{n}^i$, so I am interested in when we can find an exponent $j$ to a given exponent $i$ such that $$ A \cap \mathfrak{n}^j \subset \mathfrak{m}^i. $$ A simple example with $\bar{A} = \hat{A}$ would be $A = k[X^2]_{(X^2)}$, $\mathfrak{m} = (X^2)$, $B = k[X]_{(X)}$, $\mathfrak{n} = (X)$. Then we have $A \cap \mathfrak{n}^{2i} \subset \mathfrak{m}^i.$

Elvorfirilmathredia
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  • Please look up Artin-Rees lemma. – Mohan Jun 25 '15 at 20:11
  • I know this looks very much like Artin-Rees. The version I know [Matsumura - Commutative Ring Theory, Theorem 8.5] says $I^n M \cap N = I^{n-c}( I^c M \cap N)$. In my case I would set $M = B$, $N = A$ but I can not set $I = \mathfrak{n}$, since $I$ is required to be an ideal of $A$. – Elvorfirilmathredia Jun 25 '15 at 22:25

1 Answers1

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Artin-Rees gives you that the topology given by $\mathfrak{m}^i$ and $\mathfrak{m}^iB\cap A$ are the same. So, suffices to check that the topologies given by $\mathfrak{m}^iB\cap A$ and $\mathfrak{n}^i\cap A$ are the same. Since, as you observed, $\mathfrak{m}^iB\cap A\subset\mathfrak{n^i}\cap A$, suffices to show that for any $i$, there exists a $j$ such that $\mathfrak{n}^j\cap A\subset\mathfrak{m}^iB\cap A$. That is, $\mathfrak{n}^j\subset\mathfrak{m}^iB$. Since $B$ is a finite type module over $A$, we have $B/\mathfrak{m}^iB$ an Artin local ring with maximal ideal the image of $\mathfrak{n}$ and thus $\mathfrak{n}^j=0$ in this ring for large $j$, proving what we need.

Mohan
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