# Proof that $A/\mathfrak{p}^n \simeq A_\mathfrak{p}/\mathfrak{p}^nA_\mathfrak{p}$, for all $n$ in a Dedekind domain $A$.

$\newcommand{\p}{\mathfrak{p}}$$\newcommand{\m}{\mathfrak{m}}$$\DeclareMathOperator{im}{\mathrm{im}}$I was a bit unhappy with the proof I read which shows if $A$ is Dedekind and $\mathfrak{p}$ is a (non-zero) prime ideal then $A/\mathfrak{p}^n \simeq A_\mathfrak{p}/\mathfrak{p}^nA_\mathfrak{p}$. Can someone check my proof of the claim? I hope I'm not over simplyifying.

Consider first the natural map $\phi: A \to A_\p/\p^nA_\p$, given by $a \mapsto \ell(a) + \p^nA_\p$, where $\ell: A \to A_\p$ denotes the localization map (i.e., the canonical $a \mapsto a/1$). This map is the composition of two surjections, hence $\im(\phi) = A_\p/\p^nA_\p$. If we can show that $\ker \phi = \p^n$, then one of the isomorphism theorems tells us that $A/\p^n \simeq A_\p/\p^nA_\p$.

Now consider $\ker \phi = \{a \in A : \ell(a) \in \p^nA_\p\}$. Note that $\ell(a) \in \p^n A_\p$ if and only if $sa \in \p^n$ for some $s \not \in \p$. Hence the image of $sa$ in $A/\p^n$ is $0$.

Note however that $s \not \in \p/\p^n$, which is the unique maximal ideal in the local ring $A/\p^n$. This is because a maximal ideal in $A/\p^n$ corresponds to $\m \supset \p^n$ where $\m$ is maximal in $A$. But then $\m \supset \p$, since for $p \in \p$, $p^n \in \m$ means $p \in \m$ by induction on $n$. But also $\p$ is maximal so $\p \supset \m$, which means $\p = \m$, and thus in the quotient $\p/\p^n$.

Since $s$ is not in the unique maximal ideal of a local ring it must be a unit, and thus $a = 0$ in $A/\p^n$, which means $a \in \p^n$. Thus, $\ker \phi = \p^n$, as desired.

I was a bit pedantic above, but I hope this is correct.

• Which two surjections is that map the composition of? – MCT Feb 26 '18 at 20:45
• @MCT Let $\psi: A_\p \to A_\p/\p^n A_\p$. Isn't $\phi = \psi \circ \ell$, where $\ell$ as given above is the localization map? – Drew Brady Feb 26 '18 at 20:47
• Sure, but why is the localization map surjective? – MCT Feb 26 '18 at 20:48
• Hm. I think you have found an error in the proof. In general it is probably not surjective, since for $A \to A_\p$ to be surjective means that every $a \in A$ and any $s \not \in \p$, there is some $r \in A$ such that $rs = a$, i.e., that $A \setminus \p$ are all units, which may not hold if $A$ is not local. – Drew Brady Feb 26 '18 at 21:23
• Yea I think this looks okay! Using the fact that localization and quotients commute, one can also reduced it to showing that anything in $A/\mathfrak{p}^n$ which is not in $\mathfrak{p}$ (i.e. image is not zero under the reduction modulo $\mathfrak{p}$ map) is invertible. – MCT Feb 26 '18 at 22:32

## 1 Answer

I think your proof works (with the addendum in the comments).

One can also approach it this way: localization commutes with quotients (since localization is exact), and so it suffices to show that $A/\mathfrak{p}^n \cong (A/\mathfrak{p}^n)_{\mathfrak{p}}$. This amounts to showing that everything not in $\mathfrak{p}$ is invertible in $A/\mathfrak{p}^n$. If $s \in A \setminus \mathfrak{p}$, then $(s) + \mathfrak{p}^n = A$: if it were a strictly smaller ideal, it would be contained in a maximal ideal, which must be $\mathfrak{p}$ since $\mathfrak{p}^n + \mathfrak{q}^s = A$ in a Dedekind when $\mathfrak{p}, \mathfrak{q}$ are distinct non-zero primes. Thus the image of the ideal $(s)$ in $A/\mathfrak{p}^n$ is the whole ring, so there exists $x \in A$ such that $sx \equiv 1 \pmod {\mathfrak{p}^n}$, as desired.