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Let $R$ be an integral domain of characteristic 0 finitely generated as a ring over $\mathbb{Z}$. Can the quotient group $(R,+)/(\mathbb{Z},+)$ contain a divisible element? By a "divisible element" I mean an element $e\ne 0$ such that for every positive integer $n$ there is an element f such that $e=nf$.

As Darji points out, another way to ask the question is this: Suppose $e\in R$ has the property that for all positive integers $n$, $e$ is congruent to an integer mod $nR$. Must $e$ be an integer?

Note: I previously posted this to Math StackExchange here: http://math.stackexchange.com/questions/71031/a-question-about-the-additive-group-of-a-finitely-generated-integral-domain

TO SUMMARIZE: Qing Liu showed that in fact any non-integer rational in $R$ determines a divisible element of $(R,+)/(\mathbb{Z},+)$, and Wilberd van der Kallen showed that all divisible elements arise in this way. I wish I could accept both answers.

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Maybe a more attractive way to state the question: Can there be an element of $R$ which is equivalent to an integer in $R$ modulo $nR$ for every integer $n$, but not an integer in $R$ itself? –  darij grinberg Oct 10 '11 at 3:42
Darji, That is more attractive. Maybe that explains the no-response on MSE. On the other hand, I'm thinking that maybe there's some fact I'm not seeing about abelian groups that nails the question. –  SJR Oct 10 '11 at 3:55
I'd rather expect either a counterexample, or a proof using the Noetherianness of $R$. Possibly things like residual finitness would come into play. –  darij grinberg Oct 10 '11 at 4:04

4 Answers 4

up vote 4 down vote accepted

The answer is no in general ($e$ needs not to be in $\mathbb Z$), but one can show that $e$ is divisible in $R/\mathbb Z$ if and only if $e\in \mathbb Q\cap R$.

First let $R=\mathbb Z[1/p]$ for some prime number $p$. Then I claim that $1/p$ is divisible in $R/\mathbb Z$. Indeed for any $n\ge 1$, write $n=p^rm$ with $m$ prime to $p$. Let $a,b\in \mathbb Z$ such that $am+bp=1$. Then $$\frac{1}{p}=b+ \frac{am}{p}=b+n\frac{a}{p^{r+1}}\in \mathbb Z + nR.$$

For general $R$, denote by $D$ the elements $e\in R$ which are divisible in $R/\mathbb Z$. One can check directly that $D$ is a subring of $R$. Let us show $\mathbb Q\cap R\subseteq D$. If $e=k/q\in \mathbb Q\cap R$ with coprime $k, q$, then again using Bézout, $1/q\in R$. Then it is enough to show that $1/p\in D$ for all prime divisors $p$ of $q$. But this is done just above.

The converse is proved in Wilberd's answer ($e\in \mathbb Z[1/a]$).

Final remark: $\mathbb Q\cap R=\mathbb Z$ if and only if $\mathrm{Spec}(R)\to \mathrm{Spec}(\mathbb Z)$ is surjective. This is because the fiber of this morphism above $p$ is the spectrum of $R/pR$, and this spectrum is empty if and only if $1/p\in R$.

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Thanks for putting me straight. –  Wilberd van der Kallen Oct 10 '11 at 11:32
But you did the hardest part :) –  Qing Liu Oct 10 '11 at 12:01
I didn't check in detail, but it seems to me that your result is even true if $\mathbb{Z}$ is replaced by any PID (and, of course, $\mathbb{Q}$ by the quotient field of the PID). –  Ralph Oct 10 '11 at 12:40

As Qing Liu explains there may be such nontrivial $e$.

Suppose there was such an $e$. By Grothendieck's Generic Freeness Theorem, [Theorem 14.4 in David Eisenbud, Commutative algebra with a view toward algebraic geometry, Graduate Texts in Mathematics, Springer-Verlag, New York, 1995] there is $0\neq a\in \Bbb Z$ so that $A[1/a]$ is a free $\Bbb Z[1/a]$-module. Choose a basis and write $1$ in terms of that basis. We see $1$ lies in a direct summand spanned by finitely many basis vectors. By the structure theorem of finitely generated modules over a PID we see that in fact $Q=A[1/a]/\Bbb Z[1/a]$ is a free $Z[1/a]$-module plus a finite group. So it does not contain any nontrivial divisible element. But the image of $e$ in $Q$ is divisible. That means that $e\in \Bbb Z[1/a]$.

So far so good. The next line is wrong, as explained by Qing Liu.

But $\Bbb Z[1/a]/\Bbb Z$ does not contain any divisible element.

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Short remark: In my version of Eisenbud, Grothendieck's Generic Finiteness Theorem is Theorem 14.4. Anyway, I think that's the crucial point to reduce to the case of a finitely generated module over a PID (and thanks for letting me know Grothendieck's theorem - I didn't know that so far). –  Ralph Oct 10 '11 at 10:42
@Ralph 14.4 it is. –  Wilberd van der Kallen Oct 10 '11 at 11:05
You mean Generic <b>freeness</b>. –  Qing Liu Oct 10 '11 at 12:02
I don't understand why some people downvote this answer. It is helpful. –  Qing Liu Oct 10 '11 at 12:05
@Qing Liu Freeness it is. –  Wilberd van der Kallen Oct 10 '11 at 12:28

Edit: I misread the question and thought $R$ should be finitely generated as $\mathbb{Z}$-module (instead of as $\mathbb{Z}$-algebra). The proof below requires $R$ to be a finitely generated as $\mathbb{Z}$-module.

$R/\mathbb{Z}$ can't contain other divisible elements than $0$.

This can be seen as follows: By assumption $(R,+)$ is a finitely generated abelian group and since $R$ is integral, the group $(R,+)$ is torsion free. Thus $R$ is a finitely generated free $\mathbb{Z}$-module with $\mathbb{Z} \cdot 1_R$ as rank one sub-module. By elementary divisors there are $e_1,...,e_m \in R$ and $l \in \mathbb{Z}$ such that $R = \oplus_{i=1}^m \mathbb{Z}e_i$ and $1_R = le_1$.

Let $x= \sum_i x_ie_i \in R$ such that $\bar{x} \in R/\mathbb{Z}$ is divisible. Thus, for $n \in \mathbb{Z}$ there is $y = \sum_i y_ie_i \in R$ with $$\mathbb{Z}\cdot 1_R = \mathbb{Z}le_1 \in x-ny = (x_1 -ny_1)e_1 + ... + (x_m-ny_m)e_m.\hspace{20pt}(\ast)$$ In particular, $x_i-ny_i=0$, i.e. $n | x_i$ for $i >1$ and all $n$. Hence $x_i=0$ for $i>1$. Now choose $n:= l$. Comparing the first component in $(\ast)$ shows $l | x_1$, say $x_1= kl$. Therefore $x=k(le_1) = k \cdot 1_R$ and $\bar{x} = 0$ in $R/\mathbb{Z}$. q.e.d.

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Note that $\Bbb Z [x]$ is finitely generated as an algebra, not as an abelian group. –  Wilberd van der Kallen Oct 10 '11 at 9:51
Thanks for pointing out. –  Ralph Oct 10 '11 at 10:32

Let $e$ be an element of $\langle R,+\rangle / \langle \mathbb{Z},+\rangle$ such that $e\neq 0$.
Since the representatives of $e$ only differ in the constant term, let $m_1,...,m_n$ be non-negative integers which are not all zero such that the $(a_1)^{m_1}\cdot ... \cdot (a_n)^{m_n}$ coefficient of $e$'s representives is non-zero.
Since that coefficient is not divisible by itself plus one, $e$ is not divisible by that coefficient plus one.
Therefore the quotient group cannot contain a divisible element.

I don't know what the answer would be if you let $\; R = \mathbb{Z}[a_1,...,a_n]/I \;$ instead of $\; R = \mathbb{Z}[a_1,...,a_n] \;$.

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Ricky, Why should the representatives of $e$ differ only in the constant term? Are you assuming that the $a_i$ are algebraically independent? I am not making that assumption. Or am I missing something obvious?? –  SJR Oct 10 '11 at 3:45
The algebraic independence of the $a_i$ is basically part of the definition of $\mathbb{Z}[a_1,...,a_n]$, since the $a_i$ were not given as elements of some ring that $\mathbb{Z}$ is a subring of. $\;$ –  Ricky Demer Oct 10 '11 at 3:51
Ricky, The $a_i$ are given as elements of $R$! But anyway, I will edit my question to clarify this. Sorry for the confusion. –  SJR Oct 10 '11 at 3:57
No, the $a_i$ are used to define $R$, not given as elements of it. –  Ricky Demer Oct 10 '11 at 4:03

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