Krull dimension in quotient of local regular ring

Question

Let $(R,\mathfrak m)$ be a local ring, I an ideal of $R$ such that $R/I$ is regular.

Let $\overline{\mathfrak m}= \mathfrak m/I$ and $k:= R/\mathfrak m$.

Is it true that $\operatorname{dim}_k \mathfrak m/\mathfrak m^2 = \operatorname{dim}_k \overline{\mathfrak m}/\overline{\mathfrak m}^2 + \operatorname{dim}_k \mathfrak (m^2+I)/\mathfrak m^2$?

I don't see why this is true.

Thank you for your help.

Answer 1

I believe it is true because we have a sequence of ideals $\mathfrak{m} \supseteq \mathfrak{m}^2 +I \supseteq \mathfrak{m}^2$ whose succesive quotients are $R/\mathfrak{m}=k$-vector spaces,

and notice that $$\overline{\mathfrak{m}}/\overline{\mathfrak{m}}^2 = (\mathfrak{m}/I)/(\mathfrak{m}/I)^2 = \mathfrak{m}/(\mathfrak{m}^2 + I)$$ and then using (for example) that length is additive over filtrations the result follows.