I am now dealing with the DVR and the exact sequence. Here is one exercise I am not very clear about the structure.
Let $R$ be a DVR with maximal ideal $m$, and quotient field $K$, and suppose a field $k$ is a subring of $R$, and that the composition $k\rightarrow R\rightarrow R/m$ is an isomorphism of $k$ with $R/m$. Consider $m^n/m^{n+1}$ is an $R$-module, and so also a $k$-module, since $k\subset R$. Show that $dim_k(m^n/m^{n+1})=1$ for all $n\ge 0$.
What I am confused is that the structure of $m^n/m^{n+1}$ and how to define the dimension of $m^n/m^{n+1}$ over $R/m$, Whether $m^n/m^{n+1}$ is equal to $m^nR/m$. I am not very sure about this. What I can do to caculate this dimension. Thank you!
If $m$ is a maximal ideal of a commutative ring, then, for any $R$-module $N$, $N/mN$ is a $R/m$-module. This is because it is an $R$-module which is annihilated by the action of $m$. Very conveniently, as $R/m$ is a field, we may treat it as a vector space over $k=R/m$, and use the methods of linear algebra to study it. In particular, it has a dimension $\dim_k(N/mN)$, which is finite if $N$ is finitely generated. If $a_1,\ldots,a_r$ are generators of $N$ as an $R$-module, then their images in $N/mN$ will span it as a $k$-vector space, and so $\dim_k(N/mN)\le r$.
We are in the case $N=m^n$. Here $R$ is a DVR and $m=R\pi $ say is principal. Therefore $m^n=R\pi^n $ is also principal. So the dimension of $m^n/m^{n+1}$ is at most $1$; it could only equal zero if $m^n=m^{n+1}$. That's can't possibly be true, can it?