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Let $R=\mathbb{Z}[a_1,\ldots,a_n]$ be an integral domain finitely generated 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 $e=nf$.

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Do you mean $R$ modulo the image of $\mathbb{Z}$, or do you want to assume that $\mathbb{Z}$ injects into $R$, i.e., that $R$ has characteristic zero? – Keenan Kidwell Apr 7 '12 at 4:18

$R$ is residually finite, and so has no divisible elements since no finite ring or group $A$ has a divisible element: $nA = 0$ where $n = |A|$.

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I assume you want to write $\mathbf{Z}[x_1,\ldots,x_n]/I=R$ for your ring $R$, where $I$ is an ideal in $\mathbf{Z}[x_1,\ldots,x_n]$.

Anyway, the additive group of $R$ is not finitely generated unless $\mathrm{dim} R=0$. So you don't have any structure theorem.

It's clear that $\mathbf{Z}$ does not have any divisible elements. ($e$ is not divisible by $e+1$.) But we don't need this.

Anyway, if $R=\mathbf{Z}/m\mathbf{Z}$ ($m\geq 1$), it is clear that your quotient is the zero ring. So you can suppose that $n>1$. Now, just look at the polynomial ring $\mathbf{Z}[x_1,\ldots,x_n]$. Clearly, this doesn't have any divisible elements. If you divide out by $\mathbf{Z}$ it still doesn't have any divisible elements.

So I suspect the answer to be no.

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"the additive group of $R$ is not finitely generated unless $\dim(R)=0$." This is not true. The ring of integers in a number field is a finite $\mathbb{Z}$-module and has Krull dimension $1$. – Keenan Kidwell Apr 7 '12 at 4:17

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