4

In Proposition 10.15 in Atiyah-Macdonlad, what does the equality $\hat{\mathfrak a}=\hat A\mathfrak a$ mean?

I know that there is an isomorphism $\hat A\otimes_A\mathfrak a\cong\hat{\mathfrak a}$ and I understand what it does (what is sent to what), but I am confused as to the 'nature' of the object $\hat A\mathfrak a$. Is it the subset of elements (Cauchy sequences) of $\hat{\mathfrak a}$ of the form $(a_ia)$ where $(a_i)$ is an element of $\hat A$ (viewed as a Cauchy sequence of elements $a_i$ of $A$) and $a\in\mathfrak a$?

If so, what does $(\hat A\mathfrak a)^n$ mean? For that matter, what sort of object is $(\hat{\mathfrak a})^n$? From what I understand $\hat{\mathfrak a}$ is a module (as all completions are), so unless it is considered as an ideal of some ring, the exponent $n$ doesn't make much sense (I am pretty sure it doesn't mean direct sum of $n$ copies).

aytio
  • 584

1 Answers1

2

Suppose $R\rightarrow S$ is a ring homomorphism and $\mathfrak{a}$ is an ideal of $R$. Do you know what $\mathfrak{a}S$ means? It means the ideal in $S$ generated by the image of $\mathfrak{a}$ under the homomorphism. $\hat{A}\mathfrak{a}$ means the same thing here. And yes, $\hat{\mathfrak{a}}$ is an ideal of $\hat{A}$.

Must
  • 337
  • Thank you. Could you please clarify what $R$ and $S$ are in this context? I am guessing $R=A$ and $S=\hat A$. But then the statement $\hat{\mathfrak a}=\hat A\mathfrak a$ assumes some natural identification of $\hat{\mathfrak a}$ in $\hat A$, am I right? Also, shouldn't it be $\mathfrak a\hat A$ then? – aytio Apr 19 '14 at 04:24
  • After some more thinking, I am guessing that in the equalities $\hat{\mathfrak a}=\hat A\mathfrak a$ and $(\mathfrak a^n)^{\hat{}}=(\hat{\mathfrak a})^n$ all objects are viewed as lying in the bigger object $\hat A$ and are equalities as ideals. I was thinking of them just as modules, not as subsets of the same larger set, hence my confusion. – aytio Apr 19 '14 at 04:44
  • 1
    Since the rings are commutative $\mathfrak{a}\hat{A}=\hat{A}\mathfrak{a}$. Also, a module which is a subset of a ring is an ideal. So if you have an ideal $\mathfrak{a}$ of $A$, you have an inclusion $0\rightarrow\mathfrak{a}\rightarrow A$. Since $\hat{A}$ is flat over $A$, tensoring the previous inclusion with $\hat{A}$ will remain an inclusion and you will get $0\rightarrow\hat{\mathfrak{a}}\rightarrow\hat{A}$. So $\hat{\mathfrak{a}}$ is an $\hat{A}$-module that is a subset of $\hat{A}$, and therefore is an ideal of $\hat{A}$. – Must Apr 19 '14 at 15:29