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,

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

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 :)