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Hello, I'm reading about analytic sheaves and I've a problem to understand something that's related with commutative algebra:

Let $\mathfrak{a}\subset R$ an ideal and $M$ an $R$-module. Then,

  1. $\mathfrak{a}^n/\mathfrak{a}^{n+1}\cong \mathfrak{a}/\mathfrak{a^2}\otimes \ldots \otimes \mathfrak{a}/\mathfrak{a}^2$ (tensoring $n$-times).

  2. The sequence $0\longrightarrow \mathfrak{a}^{n+1}M\longrightarrow \mathfrak{a}^n M \longrightarrow M/\mathfrak{a}M \otimes \mathfrak{a}^n/\mathfrak{a}^{n+1}\longrightarrow 0 $ is exact.

I think that the isomorphism for the first one is just take $a_1\otimes\ldots \otimes a_n$ and take the product (which indeed belongs to the desired ideal $\mathfrak{a}^n/\mathfrak{a}^{n+1}$), but I don't see why it's injective.

For the second, I think that's true because $M/\mathfrak{a}M \cong R/\mathfrak{a}\otimes M$ so the third term is $R/\mathfrak{a}\otimes M \otimes \mathfrak{a}^n/\mathfrak{a}^{n+1} \cong M\otimes \mathfrak{a}^n/\mathfrak{a}^{n+1}$. But I don't know if it's true that the exactness of $0\longrightarrow \mathfrak{a}^{n+1}\longrightarrow \mathfrak{a}^n\longrightarrow \mathfrak{a}^n/\mathfrak{a}^{n+1}\longrightarrow 0$ implies the exactness of the desired sequence. I must to suppose that $M$ is flat? (In my context I only know that's a coherent sheaf, that's it locally finitely presented), Is $\mathfrak{a}^n/\mathfrak{a}^{n+1}$ flat as $R$-module?

Thanks a lot for your help :)

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Both are false in general. For (1), the map from left to right is surjective but has huge kernel in general: consider $\mathfrak{a} = (x,y) \subset k[x,y] = R$ for a field $k$ (in which case the left side has dimension $n+1$ over $k$ whereas the right side has dimension $2^n$ over $k$. For (2), the map from 2nd to 3rd term doesn't make sense, and using the same $R$ and $\mathfrak{a}$ and $M = \mathfrak{a}$ makes the quotient of the middle term by the left have $k$-dimension $n+2$ whereas the third term has $k$-dimension $3(n+1)$. –  user28172 Feb 25 '13 at 13:50
Typo: meant "right to left" on the first line above. –  user28172 Feb 25 '13 at 13:50
Regarding your final question, ${\mathfrak a}^n/{\mathfrak a}^{n+1}$ is pretty much never flat. Consider the exact sequence $0\rightarrow {\mathfrak a}^{n+1}\rightarrow{\mathfrak a}^n\rightarrow{\mathfrak a}^n/{\mathfrak a}^{n+1}\rightarrow 0$. If ${\mathfrak a}^n/{\mathfrak a}^{n+1}$ were flat, this would stay exact after tensoring with $R/{\mathfrak a}$. But this replaces the map ${\mathfrak a}^n\rightarrow {\mathfrak a}^n/{\mathfrak a}^{n+1}$ with an isomorphism and hence requires that you get a zero on the left, i.e. ${\mathfrak a}^{n+1}/{\mathfrak a}^{n+2}=0$, which is generally false –  Steven Landsburg Feb 25 '13 at 14:20
Thanks for your answers :) I'm going to check the text again to try to find the mistake. –  Pedro Montero Feb 25 '13 at 15:19

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