If the vector space dimension of $\mathbb{C}[[x,y]]/I$ over $\mathbb{C}$ is finite, then $I$ contains a power of $(x,y)$

Question

I am trying to understand a proof which goes like this.

If $\mathrm{dim}_{\mathbb{C}} \ \mathbb{C}[[x,y]]/I$ is finite, then $\mathbb{C}[[x,y]]/I$ has a finite composition series whose subquotients are isomorphic to $\mathbb{C}[[x,y]]/\mathfrak m$, so $\mathbb{C}[[x,y]]/I$ is annihilated by a power of $(x,y)$.

Can anyone help me understand this proof? I shall be very grateful.

Answer 1

If an $R$-module $M$ has a composition series $(0)=M_0<M_1<\dots<M_n=M$ such that $M_i/M_{i-1}\simeq R/\mathfrak m$ for all $i=1,\dots,n$ ($\mathfrak m$ is a maximal ideal of $R$), then $\mathfrak m^nM=0$.

This can be easily proven by induction on $n$. If $n=1$, then $M\simeq R/\mathfrak m$, and therefore $\mathfrak mM=0$. Now suppose that $\mathfrak m^{n-1}M_{n-1}=0$. Since $M/M_{n-1}\simeq R/\mathfrak m$ we have $\mathfrak m(M/M_{n-1})=0$, that is, $\mathfrak mM\subseteq M_{n-1}$. Then $\mathfrak m^nM\subseteq\mathfrak m^{n-1}M_{n-1}$ which implies $\mathfrak m^nM=0$.