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Let $R$ be commutative with 1.

If $M\subset R$ is maximal, $I$ is $M$-primary, then $R_M/^eI = R/I$. In particular, $R_M/^eM = R/M$ and $^eM/(^eM)^n=M/M^n$.

I am stuck in the last statement. So the second statement follows from the first directly. The first statement is proven as follows: pass to quotient $R/I$. In this quotient ring we have $\overline{M}$ is in $rad (0)$, because $I$ is $M-$primary. Then $\overline{M}\subset rad(0)\subset Jac(R/I)$. So it must take equality. Let $D=R/I - \overline{M}$ is the set of units. So $R_M/^eI = D^{-1}(R/I)= R/I$.

I don't know how the last statement follows from the above. Do we have $M/M^n=D^{-1}(M/M^n)=D^{-1}M/D^{-1}(M^n)=D^{-1}M/(D^{-1}M)^n$?

Jun Xu
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1 Answers1

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We have the exact sequences $\require{AMScd}$ \begin{CD} 0@>{}>> M/M^n@>{}>> R/M^{n}@>{}>>R/M @>{}>> 0\\ & @VVV @VVV @VVV\\ 0@>{}>>M_e/(M_e)^n @>{}>> R_M/(M_e)^n@>{}>> R_M/M_e@>{}>>0 \end{CD}

Now the 2nd and 3rd vertical arrows are equality, hence the first vertical arrow is also equality.

Evans Gambit
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