# On finite generation of a special type of Commutative ring

Let $$R$$ be a commutative ring such that for some integer $$n \ge 1$$, the ring $$\mathbb Z^n$$ injects into $$R$$ and also there is a split surjective ring homomorphism $$R \to \mathbb Z^n$$ whose kernel is the nilradical of $$R$$. So in particular, $$R \cong \mathrm{Nil}(R) \oplus \mathbb Z^n$$ as abelian groups.

My question is:

If $$R \otimes_{\mathbb Z} \mathbb Q$$ is a finite dimensional $$\mathbb Q$$-vector space, then is $$R$$ a finitely generated $$\mathbb Z$$-module ?

Note: If you are looking for a concrete example of a ring $$R$$ with the injective and split surjective conditions as mentioned ... the $$K_0$$ of any Commutative Noetherian ring is such an example ...

No. Indeed, for any abelian group $$N$$, you can make $$R=N\oplus\mathbb{Z}$$ into a ring with multiplication $$(x,a)\cdot(y,b)=(bx+ay,ab)$$ (in other words, you take formal sums of integers and elements of $$N$$ with $$N^2=0$$). This ring satisfies your hypotheses, and $$R\otimes\mathbb{Q}$$ is finite-dimensional as long as $$N\otimes\mathbb{Q}$$ is finite-dimensional. So you could take $$N$$ to be any infinite torsion group to get a counterexample.