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Let $(R,\mathfrak{m})$ be a commutative Noetherian local ring and $M$ is an $R$-module. There is an non-negative integer $t$ such that $M/(0 :_{M} \mathfrak{m}^t)$ is finitely generated. Then $$\dim M = \dim M/(0 :_{M} \mathfrak{m}^t)$$

I try to prove this proposition but i have not find out the proof. Can someone give me any idea to prove this? Thanks.

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If I understand the notation correctly, then $(0 :_{M} \mathfrak m^t)$ denotes the submodule of $M$ consisting of elements killed by $\mathfrak m^t$. Call this submodule $N$.

Consider the short exact sequence $$0 \to N \to M \to M/N \to 0.$$ If we write $I = Ann(M)$ and $J = Ann(M/N)$, then we see that $\mathfrak m^t J \subset I \subset J.$ If $M/N = 0$, so $J = R$, then $\mathfrak m^t \subset I,$ hence $R/I$ is a quotient of $R/\mathfrak m^t$, and so $R/I$ has dimension zero. Thus $\dim M := \dim M/I = 0 = \dim M/N$, as required.

If $M/N \neq 0,$ so that $J \neq R,$ then $J \subset \mathfrak m,$ and so we see that $J^{t+1} \subset I \subset J$. Since $\dim R/J^{t+1} = \dim R/J$ (the former is just a nilpotent thickening of the latter), we see that both are equal to $\dim R/I$, which is intermediate between the two, and hence again we have $\dim M := \dim R/I = \dim R/J =: \dim M/N.$ (So $M/N$ being f.g. is not actually needed.)

If instead you want to think about the dimension of the support, rather than in terms of Krull dimensions of annihilators (the two are the same in the f.g. context), then from the above exact sequence you see that $M$ and $M/N$ have identical supports, since $N$ is supported at the closed point $\mathfrak m$ (being annihilated by $\mathfrak m^t$).

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