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Let $S=k[[t^3,t^5,t^7]]$ be a formal power series over field $k$.I wanna know why $$\dim_k \operatorname{Soc}(S/t^3S)=2?$$.($\dim_k$ means dimension as $k$-vector space.)

background: $\operatorname{Soc}(M)=(0:_M m)=\lbrace x\in M ; xm=0 \rbrace$ where $(R,m)$ is a local ring and $M$ is an $R$-module.

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@YACP Yes sir :P –  rschwieb Dec 13 '12 at 15:30
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up vote 2 down vote accepted

Let $\mathfrak m=(t^3,t^5,t^7)$. It is easy to prove that $\mathfrak m^2=t^3\mathfrak m$. Then, in $\overline S=S/t^3S$, we get $\overline{\mathfrak m}^2=0$, so the socle of $\overline S$ is $\overline{\mathfrak m}$. Moreover $\overline{\mathfrak m}=(\overline t^5,\overline t^7)$ and now it's easy to see that the dimension over $k$ of $\overline{\mathfrak m}$ is $2$.

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