An example of a divisible ideal

Question

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

Answer 1

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

Answer 2

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.)

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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.)

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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.