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Let $R$ be a commutative ring (other than a field) with identity. I am looking for an example of a divisible ideal of $R$.

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What sense of divisibility? You mean it's an ideal that is a divisible group? Or it's a divisible as an $R$ module ala Lam's definition in Lectures on Modules and Rings? – rschwieb Feb 13 '13 at 18:01

Pick the ideal $I$ first. Then let $R = \mathbb{Z} \oplus I$, with multiplication $(m,i)(n,j) = (mn, mj+ni+ij)$.

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Not sure what notion of divisibility you're interested in.

If you are interested in it only being a divisible abelian group, then you can take any divisible abelian group $G$ and look at the matrix ring

$$ R=\left\{\begin{bmatrix}n&g\\0&n\end{bmatrix}\mid n\in \Bbb Z, g\in G\right\} $$

The group $G$ is embedded as an ideal consisting of the set of strictly upper triangular matrices. (Actually this is a relative construction of Hurkyl's example. This is a subring of a "split-null extension", whereas his is a Dorroh extension.)

If you are interested in the definition Thomas Andrews mentions in the comments, (where you are checking for divisibility over non-zero divisors) then my matrix ring still works, since the zero divisors are exactly the elements with $n=0$. (This version of divisibility coincides with Lam's definition below when the ring is a domain.)

There is yet another good definition in Lam's Lectures on modules and rings:

An $R$ module $M$ is divisible if for any $m\in M$ and $r\in R$ such that $r.ann(r)\subseteq ann(m)$, there exists $n\in M$ such that $m=nr$.

You can also find it proven there that a ring is von Neumann regular iff all right $R$ modules are divisible (in this sense). In particular, all of its ideals would be divisible.

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Is that definition equivalent to the one here: ? It seems related, but Lam's definition allows $r$ to be a zero-divisor as long as $m$ is annihilated by all annihilators of $r$. – Thomas Andrews Feb 13 '13 at 18:12
@ThomasAndrews It isn't the same, and Lam brings that up. He makes a case that this is a good definition in that book. I'm able to view the page in Lectures on Modules and rings on page 70, where he has a footnote addressing the definition you mentioned, so you might enjoy reading that :) – rschwieb Feb 13 '13 at 18:21
I am interested in the definition in Lam's Lectures on modules and rings when the ring is integral domain. Indeed, if $R$ is an integral domain (commutative ring) and $M$ is an $R$-module. Then it is said that $M$ is divisible provided that $RaM=M$ for every $a\in R$. Now, I have an example of a divisible ideal of $R$ as an $R$-module. – user62181 Feb 14 '13 at 5:57
@user62181 Oh, I just saw why it was obvious a nontrivial ideal of a domain can't be divisible. Let $a$ be nonzero in $I$. In particular for the choice of $a=r$, there must be a $b\in I$ such that $ab=a$, but this means that $a(b-1)=0$, and so $b=1$, a contradiction to $I$ being nontrivial. – rschwieb Feb 14 '13 at 16:02
@user62181 And if one hoped that a domain $R$ were divisible as a module over itself, then given any nonzero $a,b$ one could find $c$ such that $a=cb$, and in turn one could find $d$ such that $b=da$. But then $a=cda$ implies $cd=1$, and so every pair of nonzero elements are associates, but that means that $R$ is a field. There just aren't any divisible ideals in nonfield domains. – rschwieb Feb 14 '13 at 18:58

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