3
$\begingroup$

I was wondering whether $\mathbb{Q}/\mathbb{Z}$ can be decomposed into a direct sum of its subgroups and I thought of the following decomposition: $$\mathbb{Q}/\mathbb{Z} = \bigoplus_{p \text{ prime}}\left(\mathbb{Z}\left[\frac{1}{p}\right]\Big/\mathbb{Z}\right).$$

The isomorphism is quite obvious as via CRT any fraction $\frac{m}{p_1^{\alpha_1}...p_k^{\alpha_k}}$ can be uniquely written as a sum $\frac{m_1}{p_1^{\alpha_1}} + ... + \frac{m_k}{p_1^{\alpha_k}} \:\mathrm{mod}\,\mathbb{Z}$. Firstly, I want to ask: is this reasoning correct and this is indeed how $\mathbb{Q}/\mathbb{Z}$ is decomposed?

Afterwards, I got a feeling that this somehow seems like something more general and not linked to this particular example. I figured that each summand in the decomposition above is in fact a localization of $\mathbb{Q}/\mathbb{Z}$ at a prime ideal $(p) \subset \mathbb{Z}$. So, if we consider $M = \mathbb{Q}/\mathbb{Z}$ as an $R$-module where $R = \mathbb{Z}$, we get the following decomposition: \begin{align}M = \bigoplus_{\mathfrak{p} \in \mathrm{Spec}R \setminus \{(0)\}} M_\mathfrak{p}.\end{align}

So, my question is quite vague but I am too curious not to ask: $$\textbf{under which conditions can a module be decomposed in the above fashion?}$$

For finitely generated modules over principal ideal domains to me this feels very much Structure Theorem-ish. That is, we know that if $M$ is a f.g. module over a principal domain $R$ it can be decomposed as $$M = R^{\oplus s} \oplus R/(p_1^{\alpha_{11}}) \oplus ... \oplus R/(p_1^{\alpha_{1r_1}}) \oplus ... \oplus R/(p_k^{\alpha_{k1}}) \oplus ... \oplus R/(p_k^{\alpha_{kr_k}}).$$

Then the localization of $M$ at a prime ideal $(p_i)$ is $R_{(p_i)}^{\oplus s} \oplus R_{(p_i)}/(p_i^{\alpha_{i1}})R_{(p_i)} \oplus ... \oplus R_{(p_i)}/(p_i^{\alpha_{ir_i}})R_{(p_i)}$.

So to me it seems that the necessary condition for the decomposition to work is

1. that the module is equal to its torsion submodule;

also it seems that we also have to require that

2. for each prime ideal $\mathfrak{p}$ in $R$ and each nonnegative integer $\alpha$ we have equality $R/\mathfrak{p}^\alpha = R_\mathfrak{p}/\mathfrak{p}^\alpha R_\mathfrak{p}$.

Under which conditions does the second assumption hold? Will this two conditions be sufficient for a desired decomposition to occur, not necessarily if $M$ is f.g. and $R$ is a PID? Under which conditions such decomposition is possible?

I would really appreciate any help or advice. Thank you in advance!

And I apologize if my question is flawed in some way. I tried to express my understanding of the problem as much as possible and I hope I haven't missed anything.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Your condition $\textbf{2}$: $R/\mathfrak{p}^\alpha = R_\mathfrak{p}/\mathfrak{p}^\alpha R_\mathfrak{p}$ generally holds when $\mathfrak{p}$ is a maximal ideal of $R$. For if $r \text{ mod } \mathfrak{p}^\alpha$ is in the kernel of the natural map $R/\mathfrak{p}^\alpha \to R_\mathfrak{p}/\mathfrak{p}^\alpha R_\mathfrak{p}$, we have $sr \in \mathfrak{p}^\alpha$ for some $s \in R \setminus{\mathfrak{p}}$. Then $R = Rs+\mathfrak{p}^\alpha$, say $1 = us+v$ for suitable $u \in R$ and $v \in \mathfrak{p}^\alpha$, so that $r=usr+vr \in \mathfrak{p}^\alpha$. That is, $r \text{ mod } \mathfrak{p}^\alpha = 0$. And every $s \in R \setminus{\mathfrak{p}}$ is already invertible in $R/\mathfrak{p}^\alpha$, because $Rs+\mathfrak{p}^\alpha = R$. Hence $R/\mathfrak{p}^\alpha \twoheadrightarrow R_\mathfrak{p}/\mathfrak{p}^\alpha R_\mathfrak{p}$.

This may fail for prime ideals that are not maximal. E.g., let $R = k[X,Y,Z]/(XY-Y^2-Z^2)$, where $k$ is a field, and $\mathfrak{p} = (\overline{Y},\overline{Z})$. Then $\overline{X}\overline{Y} \in \mathfrak{p}^2$ while $\overline{X} \notin \mathfrak{p}$ and $\overline{Y} \in \mathfrak{p} \setminus{\mathfrak{p}^2}$. Thus $\overline{Y} \text{ mod } \mathfrak{p}^2$ becomes zero when localizing at $\mathfrak{p}$.

The natural map $f:M\to\bigoplus_{\mathfrak{p} \in \mathrm{Max}(R)} M_\mathfrak{p}$ is well defined iff each $x\in M$ is zero in $M_\mathfrak{p}$ for almost all maximal ideals $\mathfrak{p}$. That is, iff the annihilator $\mathrm{ann}_R(x)$ of each $x$ is contained in only finitely many maximal ideals of $R$.

If $f$ is well defined, it is injective, but not usually surjective. It is in the case of a finitely generated torsion module $M$ over a principal ideal domain $R$, for the reasons you give. But, given any commutative ring $R$, the map $f$ will be an isomorphism for any $M$ of the form $\bigoplus_{\mathfrak{p} \in \mathrm{Max}(R)} N(\mathfrak{p})$, when each $N(\mathfrak{p})$ is an $R_{\mathfrak{p}}$-module with $R$-annihilator radical $\sqrt{\mathrm{ann}_R(N(\mathfrak{p}))} = \mathfrak{p}$. And such $M$ need not be f.g. or torsion, of course.

$\mathbb{Z}\left[\frac{1}{p}\right]=\mathbb{Z}_p=S^{-1}\mathbb{Z}$, where $S$ is the multiplicative system consisting of the powers of $p$. As $\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}$ is a proper subset of $\mathbb{Q}/\mathbb{Z}$, it is therefore $\textbf{not}$ a localization of $\mathbb{Q}/\mathbb{Z}$. Indeed, $\mathbb{Q}/\mathbb{Z}$ is a divisible $\mathbb{Z}$-module, so it is equal to each of its localizations (ignoring the degenerate localization $0=S^{-1}\mathbb{Q}/\mathbb{Z}$, where $S=\{0\}$). Your decomposition of $\mathbb{Q}/\mathbb{Z}$ is correct, but not of the form discussed above.

$\endgroup$
3
  • $\begingroup$ Thank you! I guess, in case $\mathfrak{p}$ is maximal, we could argue that the ring $R / \mathfrak{p}^\alpha$ is local with max. ideal $\mathfrak{p} / \mathfrak{p}^\alpha$, thus it satisfies the UMP of localization, 'cause its invertible elements are precisely the ones not in $\mathfrak{p}$. Is that right? So I believe it makes sense to pose my question in context of one-dimesional domains, yes? $\endgroup$ Commented May 30, 2022 at 10:28
  • $\begingroup$ I'm curious about a condition that $\mathrm{ann}_R(x)$ is contained in only finitely many max. ideals. Are there any cases when this condition is satisfied automatically? What about one-dimesional domains in particular? $\endgroup$ Commented May 30, 2022 at 10:37
  • 1
    $\begingroup$ This trivially holds if $R$ is semi-local (finitely many maximal ideals). And for Dedekind domains (assuming $\text{ann}_R(x) \ne 0$, of course). These are one-dimensional. $\endgroup$ Commented May 30, 2022 at 11:57

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.