Ideals of the unitalization

Question

We know that every commutative ring can be embedded in a ring with identity as follow:

Let $R$ be ring and $R_1=R\times \mathbb{Z}=\{(r,n)\mid r\in R,n\in \mathbb Z\}$. This is a ring with addition defined as $$(r,n)+(s,m)=(r+s,n+m)$$ and multiplication defined as $$(r,n)\cdot(s,m)=(rs+ns+mr,nm).$$ This ring has unity as $(0,1)$. Now we can easily show that if we define an homomorphism from $R\to R_1$ as $f(r)=(r,0)$ $\forall r\in R$ then $R \cong f(R)\subseteq R_1$. Hence $R$ is embeddable in $R_1$ which has unity $(0,1)$.

Now I want to know that if there is any description for the ideals of $R \times \mathbb{Z} $, or special prime or maximal ideals.

Answer 1

If $I$ is an ideal of $R$, consider $I\times \mathbb{Z}$. You have that for any element $(r,n) \in R \times \mathbb{Z}$, for any $(i,m) \in I \times \mathbb{Z}$ $$(r,n)(i,m) = (ri+ni+mr,mn)$$ which a priori is not an element of $I \times \mathbb{Z}$, unless $m=0$. So, for sure, any ideal of $R$ becomes an ideal of $R \times \mathbb{Z}$ via $f$ (including $R$ itself)

There is no ideal of the form $(0,p)$ for any $p$, since multiplication by $(r,n)$ makes the first factor a multiple of $r$.

However, consider an ideal $J$ of $\mathbb{Z}$ (they are of the form $n\mathbb{Z}$). Then $R \times J$ is an ideal of $R \times \mathbb{Z}$ since $$(r,n)(s,j) = (rs+ns+jr,jn)$$

Answer 2

I don't think I've seen a proof that lists exhaustively the ideals of this construction (the Dorroh extension). People have certainly explored the problem though. A more recent work is this one by Mesyan which gets some good results in certain conditions (see sections 6-8).

Here are some basics to take away:

1. if $R$ has identity then $R_1$ is isomorphic to $R\times \mathbb Z$ and finding the prime and maximal ideals is trivial.
2. if $R$ does not have identity things are not so easy, but at least you know the prime and maximal ideals containing $R\times\mathbb Z$, since they correspond to those in $\mathbb Z\cong R_1/R\times\{0\}$.
3. At least the nilradical of $R_1$ is easy to find for commutative rings: it's just $nil(R)\times \{0\}\subseteq R_1$.
4. any idempotents of $R_1$ are of the form $(e, 0)$ with $e^2=e$ or $(f, 1)$ with $f^2=-f$.