# Isomorphism of quotients of powers of maximal ideals

Let $R$ be an integral domain, and $\mathfrak{m}$ a maximal ideal of $R$. Let $R_\mathfrak{m}$ denote the ring localized at $\mathfrak{m}$, and let $\mathfrak{m}_\mathfrak{m} = \mathfrak{m}R_\mathfrak{m}$ denote the maximal ideal of $R_\mathfrak{m}$.

Then we have an isomorphism $$R/\mathfrak{m}^n \cong R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^n,$$ for $n \geq 1$. (See for example Neukirch, Algebraic Number Theory, pg. 66).

My question:

Can this result extend to an isomorphism $$\mathfrak{m}^k/\mathfrak{m}^n \cong \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n,$$ for $1 \leq k \leq n$? It would seem the proof goes through, but I wanted to check with the experts, since I hadn't seen this result explicitly written down before.

2nd question (added later):

Is the "integral domain" condition really necessary to either Neukirch's result or the other one I suggested?

Yes, $\mathfrak{m}^k/\mathfrak{m}^n \cong \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$.

This immediately results from the commutative diagram $$\begin{array} 00 & \longrightarrow \mathfrak{m}^k/\mathfrak{m}^n\longrightarrow & R/\mathfrak{m}^n & \longrightarrow & R/\mathfrak{m}^k& \longrightarrow 0\\ & \quad \quad \downarrow & \downarrow \cong & & \downarrow \cong\\ 0 &\longrightarrow \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n\longrightarrow & R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^n & \longrightarrow & R_\mathfrak{m}/\mathfrak{m}_\mathfrak{m}^k& \longrightarrow 0 \end{array}$$ Since you know that the the two rightmost vertical maps are isomorphisms, the leftmost vertical map is also an isomorphism, by the snake lemma (say).

• Thanks! That's very nice. Aug 28, 2013 at 22:08
• You are welcome, John. Aug 28, 2013 at 22:10

I suggest a different approach in order to prove that $\mathfrak{m}^k/\mathfrak{m}^n \simeq \mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$ which also contains the case $k=0$.

Obviously $\mathfrak{m}_\mathfrak{m}^k/\mathfrak{m}_\mathfrak{m}^n$ is isomorphic to $S^{-1}(\mathfrak{m}^k/\mathfrak{m}^n)$, where $S=R-\mathfrak m$. If we set $M=\mathfrak{m}^k/\mathfrak{m}^n$ we have to prove that $M\simeq S^{-1}M$. But $\mathfrak m^tM=0$ for $t=n-k\ge 1$, and thus $M$ is in fact an $\overline R=R/\mathfrak m^t$-module. Now we can replace $S$ by $\overline S=\overline R-\overline{\mathfrak m}$ and let us see what we have now:

$\overline R$ is a local ring with maximal ideal $\overline{\mathfrak m}$, $\overline S=\overline R-\overline{\mathfrak m}$, and $M$ an $\overline R$-module. We want to prove that $M\simeq\overline{S}^{-1}M$.

But there is nothing to prove once we note that $\overline S=U(\overline R)$.

• Thanks. Perhaps you could elaborate a bit on why $M\simeq\overline{S}^{-1}M$ implies $M\simeq S^{-1}M$? Is that automatic? Or is there something to show there? Aug 29, 2013 at 16:33
• @JohnM $S^{-1}M$ is isomorphic to $\overline{S}^{-1}M$ canonically, that is, by sending $x/s$ to $x/\overline s$. (Take into account that $\overline ax:=ax$.)
– user26857
Aug 29, 2013 at 16:36

Well I found this result in Milne's Commutative Algebra notes, Prop 5.8, and the ring need not be either Noetherian or an integral domain.

• Nevertheless, thanks for your answer and posting the diagram! Aug 29, 2013 at 0:14