Uniform dimension of completions

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Let $R$ be a filtered ring, and let $M$, $N$ be filtered $R$-modules such that $N\subseteq M$, and $N$ is dense in $M$.

Is it true that in this case, the uniform dimension of $M$ is equal to the uniform dimension of $N$, and if not, can you think of a counterexample?

Do you know any conditions on the filtration that will allow this result to hold?