Does this ring constructed from a group have an accepted name? Let $(G, +, 0)$ be a commutative group. Consider the set $R = G \times \mathbb{Z}$, equipped with the following operations:
\begin{align}
(g, m) + (h, n) &= (g + h, m + n) \\
(g, m)(h, n) &= (g + mh, mn)
\end{align}
It is easy to see that $R$ is a ring (in general non commutative) in which $(0, 0)$ is the zero and $(0,1)$ is the unit.
Question. This construction is likely to be standard. Does it have an accepted name?
Remark. I am not asking for a proof that $R$ is a ring, just a reference to the appropriate terminology.
EDIT. The answer by rschwieb shows that $R$, as defined, is actually not a ring. He also gives the right definition as well as the appropriate terminology.
 A: It does not form a ring because it does not distribute properly:
$(a,b)((c,d)+(e,f))=(a,b)(c+e,d+f)=(a+bc+be, bd+bf)$
$(a,b)(c,d)+(a,b)(e,f)=(a+bc,bd)+(a+be, bf)=(2a+bc+be, bd+bf)$
But if you mean $(g,m)(h,n)=(g\boldsymbol{n}+mh, nm)$, then that structure is called the "trivial extension" or "split null extension".
A: With the correct definition given by rschwieb, one context for this might be the following. 
Suppose that $(S, +, \cdot)$ is a ring, possibly without unity. Define $R = S \times \mathbb{Z}$ with the following operations:$$(s, m) + (s', m') = (s+s', m+m')$$
$$(s,m) \cdot (s',m') = (ss' + ms' + m's, mm')$$
Then $R$ is a ring with zero $0 = (0,0)$ and multiplicative identity $1 = (0,1)$. Furthermore, we have an embedding of $S$ into $R$ via $s \mapsto (s,0)$.
I am not an expert, but I think in the literature $R$ here is called the Dorroh extension of $S$. The main point of this construction is that it gives you a way to embed a ring without unity into a ring with unity.
In your case, the commutative group $(G, +, 0)$ becomes a ring without unity when we define the multiplication by $gg' = 0$ for all $g,g' \in G$. Doing the construction above for $G$ we then end up with the ring $R = G \times \mathbb{Z}$ which has multiplication defined as in the answer by rschwieb.
