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In the proof of the fact that "if $I$ is an ideal of the regular local ring $(R,m)$ such that $R/I$ is regular then $I$ can be generated by part of a minimal generating set of of $m$", I saw in a textbook that the author takes the dimension of regular ring $R/I$ equal to $t$. The projections $m→m ̅→m ̅/m ̅^2$ (of $R$-modules) induce a surjective homomorphism $f:m/m^2→m ̅/m ̅^2$ of $R/m$-vector spaces, with $\ker f=(I+m^2)/m^2$. Hence the dimension (over $R/m$) of $(I+m^2)/m^2$ is $d-t$. ... My questions are:

(1) What is the formula of $f$?

(2) Why the dim of $(I+m^2)/m^2$ is $d-t$?

Thanks in advance.

user26857
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karparvar
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1 Answers1

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I guess $\overline m=m/I$. Then $\overline m^2=(m^2+I)/I$ and $\overline m/\overline m^2=m/(m^2+I)$.

The formula for $f$ is pretty clear: $f(x\bmod m^2)=x\bmod (m^2+I)$.

From the short exact sequence $$0\to(m^2+I)/m^2\to m/m^2\to m/(m^2+I)\to 0$$ you get what you want (I suppose that $d=\dim R$, that is, $d=\dim_{R/m}m/m^2$).

user26857
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