I am wondering about a question:

We know that $\mathbb{Q}/\mathbb{Z}$ is torsion group and $\mathbb{Q}/\mathbb{Z}=\bigoplus_{p\text{ prime}}\mathbb{Q}/\mathbb{Z}(p)$ where $\mathbb{Q}/\mathbb{Z}(p)=\{\bar{\frac{a}{p^{\alpha}}\mid a\wedge p=1 \text{ and } \alpha\in\mathbb{N}}\}$

Can we generalize this result as follows?

Let $A$ be a ring. Consider $\operatorname{Frac}(A)/A$ as $A$-module, I proved that $\operatorname{Frac}(A)/A$ is a torsion module like we do in $\mathbb{Q}/\mathbb{Z}$. I want to prove that is direct sum of something which is a generalization of primes. I think about the coprime ideal, but I can not found a similar result.

My question is: is this generalization valid or not, and if yes can we construct submodules that give the whole module via the direct sum?

  • 1
    $\begingroup$ $Frac(A)/A$ will always be a torsion $A$-module, so this assumption is redundant. $\endgroup$
    – Daniel W.
    Commented Apr 13, 2019 at 6:35
  • 2
    $\begingroup$ In the case where $A$ is a Bézout domain that is a UFD (so in particular if $A$ is Euclidian), this will definitely work. Just take a system of representatives of prime elements up to associates (instead of the prime numbers in your example). $\endgroup$
    – Daniel W.
    Commented Apr 13, 2019 at 6:39
  • 3
    $\begingroup$ If $A=k[X,Y]$, polynomial ring of 2 variables over a field, I don't know if $Frac(A)/A$ is a direct sum of two nonzero $A$-modules. $\endgroup$
    – YCor
    Commented Apr 13, 2019 at 6:53
  • $\begingroup$ but can it work for special case of ring other than euclidian? can you write it down $\endgroup$
    – Mary Maths
    Commented Apr 13, 2019 at 9:32

1 Answer 1


Definitions and Remarks: Let $D$ be an integral domain.

  1. $D$ is called a Bézout domain if it satisfies one of the following two equivalent conditions:

(a) Every finitely generated ideal of $D$ is a principal ideal.

(b) For all $d_1,...,d_n \in D$ with $ d = \gcd(d_1,...,d_n)$ there exist $a_1,...,a_n \in D$ such that $d = a_1d_1 + ... + a_nd_n$.

  1. $D$ is called atomic if every non-zero element $a \in D$ (that is not a unit) has a finite decomposition $a = a_1...a_n$ into irreducible elements $a_1,...,a_n \in D$. If $P$ is a full system of representatives of irreducible elements in $D$ with respect to associates, then every $a \in D$ has a decomposition $a = \varepsilon p_1^{e_1}...p_n^{e_n}$ where $\varepsilon \in D^\times$, $p_i \in P$ and $e_i \in \mathbb{N}$.
  2. If $D$ is a unique factorization domain, then $D$ is atomic and every irreducible element of $D$ is prime. Moreover, the decomposition in (2) is unique up to reordering.

In the following $D$ denotes an integral domain with quotient field $K$. $P \subseteq D$ is a full system of representative of irreducible elements in $D$ with respect to associates. \ For every $p \in P$, we denote $D_{[p]} = \{ \frac{a}{p^n} | a \in D, n \in \mathbb{N}_0 \}$. $D_{[p]}$ is the localization of $D$ at the multiplicatively closed subset $[p] = \{ p^n | n \in \mathbb{N}_0 \}$ and is therefore a subring of $K$ containing $D$, and in particular a $D$-submodule of $K$.

Lemma: Let $D$ be a unique factorization domain and $p_1,...,p_n \in P$ such that $p_i \neq p_j$ for $i \neq j$. Then for all $a_1,...,a_n \in D$ with $\gcd(a_i,p_i) = 1$ and $e_1,...,e_n \in \mathbb{N}_0$ such that $\frac{a_1}{p_1^{e_1}} + ... + \frac{a_n}{p_n^{e_n}} \in D$ it follows that $e_1 = ... = e_n = 0$. In particular, we have a direct sum $\bigoplus_{p \in P}(D_{[p]}/D)$.

Proof: Let $ \frac{a_1}{p_1^{e_1}} + ... + \frac{a_n}{p_n^{e_n}} = d \in D$. Then we have $a_1 \prod_{i \neq 1} p_i^{e_i} = d \prod p_i^{e_i} - a_2 \prod_{i \neq 2} p_i^{e_i} - ... - a_n \prod_{i \neq n} p_i^{e_i}$. As the right hand side is divisible by $p_1^{e_1}$, so is the left hand side. But since the decomposition into prime elements is unique in a UFD and we assumed $\gcd(a_1,p_1)$ to be $1$, the left hand side is not divisible by $p_1$, which implies $e_1 = 0$. Proceed analogously for $e_2 = ... = e_n = 0$.

Lemma: Let $D$ be a Bézout domain that is also a unique factorization domain. Then every $k \in K$ has a decomposition $k = \frac{a_1}{p_1^{e_1}} + ... + \frac{a_n}{p_n^{e_n}}$, where $a_i \in D$, $p_i \in P$ and $e_i \in \mathbb{N}_0$. In particular $K = \sum_{p \in P}D_{[p]}$ and $K/D = \sum_{p \in P}(D_{[p]}/D)$.

Proof: Let $k = \frac{a}{b} \in K$, where $a \in D$ and $b \in D \setminus \{ 0 \}$. Then $b$ has a decomposition $b = \varepsilon p_1^{e_1} ... p_n^{e_n}$, where $\varepsilon \in D^\times$, $e_i \in \mathbb{N}$ and $p_i \in P$. Since $D$ is a unique factorization domain $\gcd(\prod_{i \neq 1} p_i, ... ,\prod_{i \neq n} p_i) = 1$. Since $D$ is Bézout, there exist $ d_1,...,d_n \in D$ such that $1 = d_1 \prod_{i \neq 1} p_i + ... + d_n \prod_{i \neq n} p_i$. So we have $\varepsilon^{-1}a = \varepsilon^{-1} a d_1 \prod_{i \neq 1} p_i + ... + \varepsilon^{-1} a d_n \prod_{i \neq n} p_i$. If we now set $a_i = \varepsilon^{-1} a d_i$, this easily leads to $\frac{a_1}{p_1^{e_1}} + ... + \frac{a_n}{p_n^{e_n}} = \frac{a}{b}$.

Corollary: Let $D$ be a Bézout domain that is also a unique factorization domain. Then $K/D = \bigoplus_{p \in P}(D_{[p]}/D)$.

Finally, in order to see that the assumptions in the Lemmata are strictly weaker than Euclidian and that they are not redundant (at least for the proof I gave), I want to give some examples:

  1. $\mathbb{Z}[\sqrt{-19}]$ is a PID (so UFD and Bézout) but not Euclidean.
  2. The ring of all rational polynomials with integer constant term $\mathbb{Z} + x \mathbb{Q}[x]$ is Bézout but not a PID.
  3. $\mathbb{Z}[x]$ is a UFD but not Bézout.

Maybe there is a chance to a generalization of these observations to Krull domains or Prüfer domains, the first being (in some sense) close to UFDs, as the second to Bézout domains. But since a non-Noetherian Prüfer domain can never be Krull (and one may wants to use the good properties of both classes of rings), it could be fruitfull to try it for Dedekind domains.


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