Why $\mathcal{O}_{X,x} \cong \oplus_i \mathfrak{m}^{i}/\mathfrak{m}^{i+1}$ as a $k$-vector spaces under some conditions? I'm reading the Gortz's Algebraic Geometry, p.327 and some question arises.
In the page, he argues as follows :
Let $k$ be a field and $X=\operatorname{Spec}A$ be a finite $k$-scheme. Then $X$ satisfies the equivalent properties of 5.20.

Thus we have $A= \prod_{x\in X}\mathcal{O}_{X,x}$, where $\mathcal{O}_{X,x}$ is a local finite-dimensional $k$-algebra.
Let us denote by $\mathfrak{m}$ the maximal ideal of $\mathcal{O}_{X,x}$. Then we have
$$ \mathcal{O}_{X,x} \cong \oplus_i \mathfrak{m}^{i}/\mathfrak{m}^{i+1}$$
as $k$-vector spaces.
Q.1 And why this isomorphism is true? My first attempt is to show that $\operatorname{dim}_k\oplus_i \mathfrak{m}^{i}/\mathfrak{m}^{i+1} = \operatorname{dm}_k\mathcal{O}_{X,x}$ And I stuck at some part of calculation of the dimension. Note that since $\mathcal{O}_{X,x}$ is artinian ( $ \because \operatorname{dim}X=0$), $\mathfrak{m} \supseteq \mathfrak{m}^{2} \supseteq \cdots$ stabilizes so that there exists $n$ such that $\mathcal{O}_{X,x} \supseteq \mathfrak{m} \supseteq \mathfrak{m}^{2} \supseteq \cdots \mathfrak{m}^{n} \supseteq \mathfrak{m}^{n+1} = \mathfrak{m}^{n} = \cdots$.
So, note that
$$\operatorname{dim}_{k}\oplus_i \mathfrak{m}^{i}/\mathfrak{m}^{i+1} = \operatorname{dim}_k(\mathcal{O}_{X,x}/\mathfrak{m}) + \operatorname{dim}_k(\mathfrak{m}/\mathfrak{m}^{2}) + \cdots + \operatorname{dim}_k (\mathfrak{m}^{n-1}/\mathfrak{m}^{n}) + \operatorname{dim}_k(\mathfrak{m}^{n}/\mathfrak{m}^{n+1})(=0) + 0 + \cdots = \operatorname{dim}_k\mathcal{O}_{X,x} - \operatorname{dim}_k\mathfrak{m}^{n}$$
(True?)
And issue that annoying me is the appearence of the term $\operatorname{dim}_k\mathfrak{m}^{n}$. Can any one rescue from this embarrassing situation?
EDIT : I think that we may use the Nakayama Lemma to guarantee that $\mathfrak{m}^{n}= 0 $. Does it works?
 A: The fact that $\mathfrak{m}^n = 0$ combines both Nakayama's Lemma together with topology. The ring $\mathcal{O}_{X,x}$ is local, of finite-type, defined over $k$, so it is a local Noetherian ring. In such a ring one necessary has that $\bigcap_n  \mathfrak{m}^n = 0$, and since it is Artinian we can conclude that some power of $\mathfrak{m}$ must be equal to zero.
So why does $\bigcap_n \mathfrak{m}^n = 0$? If we regard $\mathcal{O}_{X,x}$ as a module over itself then we can define the $\mathfrak{m}$-topology. The "Artin-Rees Lemma" says (reallly a consequence of this lemma), that the subspace topology on $\mathfrak{m}$ and the $\mathfrak{m}$-topology on it are the same. If we are willing to accept that the two natural topologies on this subspace $\mathfrak{m}$ are equal then $I = \bigcap_n \mathfrak{m}^n$ is the intersection of all neighborhoods of $0$ in the $\mathfrak{m}$-topology. However, $\mathfrak{m}I$ is also open, since the two topologies are equal, and we conclude that $\mathfrak{m}I = I$, and now by Nakayama's Lemma we can conclude that $I = 0$.
