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On page 109 of Atiyah-Macdonald, the authors let $A$ be a Noetherian ring with an ideal $\mathfrak{a}$. We use the notation $\hat A$ for the $\mathfrak{a}$-dic completion of $A$. They say that we have, $$ \hat {\mathfrak{a}} = \mathfrak{a}\hat A $$

I am trying to understand the use of this notation. First of all, does $\hat {\mathfrak{a}}$ denote the completion of $\mathfrak{a}$ as a module over $A$? Since $\mathfrak{a}$ is not a ring, the only kind of completion that can make sense is the module completion.

My second question is that, can be say that $\hat {\mathfrak{a}}\subseteq \hat A$? This is because, $$ \hat {\mathfrak{a}} = \lim_{\leftarrow} \mathfrak{a}/\mathfrak{a}^n\subseteq \lim_{\leftarrow} A/\mathfrak{a}^n$$

My third question is that, is $\mathfrak{a}\hat A$ (in the book they switch the order, and write, $\hat A \mathfrak{a}$), the notation for the "ideal-module product" i.e. given an $A$-module $M$ with ideal $I$, we denote $IM = \{ \sum xm | x\in I, m\in M\}$?

My fourth question is, when the authors write in the book, $$ \hat A \mathfrak{a}^n = \left( \hat A \mathfrak{a} \right)^n $$ what does the expression on the RHS even mean? What is the proper way of interpreting that product.

I think I understand, but I just want to double-check on the notation.

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The answer to your first three questions is yes.

For your fourth question, the equation simply says that the ideal of $\hat{A}$ generated by the $n$-th power of the ideal $\mathfrak{a}$ of $A$ over $\hat{A}$ (this is the LHS) is the $n$-th power of the ideal of $\hat{A}$ generated by $\mathfrak{a}$ over $\hat{A}$ (RHS). More explicitly, both $\hat{A} \mathfrak{a}^n$ and $\hat{A} \mathfrak{a}$ are ideals of $\hat{A}$. Then the equation says that the ideal $\hat{A} \mathfrak{a}^n$ is equal to the product ideal $(\hat{A} \mathfrak{a})^n$.

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  • $\begingroup$ For my fourth question, I was thinking of those products as products of algebras. If we are given $M$, which is an $A$-algebra, and two subalgebras, then we can define the product of two subalgebras in a similar fashion as a product of two ideals. After all, $\hat A$ and $\mathfrak{a}$ are $A$-algebras. (Yes, $\mathfrak{a}$ is not a ring, so by definition it is not an $A$-algebra, but as a module, its multiplication associates with module multiplication, and that is all we need.) $\endgroup$ – Nicolas Bourbaki Jan 10 '15 at 17:47

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