Identify the ring $\mathbb{Z}[x]/(2x)$ 
Consider the quotient ring $R:=\mathbb{Z}[x]/(2x)$. Could somebody help me to identify $R$ with some well-known ring? 

Thank You.
 A: Consider the two natural ring homomorphisms $\mathbb{F}_2[x] \to \mathbb{F}_2$, $x \mapsto 0$ and $\mathbb{Z} \to \mathbb{F}_2$ and let $P = \mathbb{F}_2[x] \times_{\mathbb{F}_2} \mathbb{Z}$ be their fiber product. The element $(x,0)$ satisfies $2(x,0)=0$. It generates $P$: If $(f,z) \in P$, i.e. $f \in \mathbb{F}_2[x]$ and $z \in \mathbb{Z}$ with $f(0) = z \bmod 2$, then one checks $(f,z)=g((x,0))$ for any lift $g \in \mathbb{Z}[x]$ of $f$.  It follows that there is a surjective homomorphism of rings $\mathbb{Z}[x]/(2x) \to P$ which maps $[f]$ to $(x,1)$. It is not hard to show that it is also injective, and therefore an isomorphism.
Geometrically, $\mathrm{Spec}(R)\cong \mathrm{Spec}(P)$ is the gluing of $\mathrm{Spec}(\mathbb{F}_2[x])$ and $\mathrm{Spec}(\mathbb{Z})$ along the closed points $(x) \sim (2)$. You get a more classical example by replacing $\mathbb{Z}$ by $\mathbb{F}_2[y]$ and the point $(2)$ by $(y)$, because then we glue two affine lines over $\mathbb{F}_2$ along their origins, which results in the union of the two coordinate axes, which has the coordinate ring $\mathbb{F}_2[x,y]/(xy)$.  See Karl Schwede's paper Gluing schemes and a scheme without closed points for more about these gluings.
A: $\mathbb{Z}[x]/(2x)$ has as its additive group $\mathbb{Z}\oplus x\mathbb{Z}_2[x]$ so every element is an integer $a$ plus a polynomial in $\mathbb{Z}_2[x]$ with zero constant term. $$(a+p(x))(b+q(x))=ab+[b]_2p(x)+[a]_2q(x)+p(x)q(x)$$
A: Let me add a basic comment that might help some people to understand what is the structure of the elements of these rings without going into advanced mathematics:
Let $\sum a_i X^i \in \mathbb{Z}[X]$ then divide coefficients $a_i$ with $i \ge 1$ by 2 to obtain:
$a_i = (2q_i+r_i) = 2q_i+r_i$
$a_iX^i = (2q_i+r_i)X^i = 2q_iX^i+r_iX^i$
Then you can write the original polynomial as:
$(a_0 + \sum r_i X^i) + (\sum 2q_iX^i)$ where $r_i \in \mathbb{Z}_2, q_i \in \mathbb{Z}$
In the quotient, the right part disappears since it is contained in the ideal $\langle 2X \rangle$ and therefore you obtain:
$a_0 + \sum r_i X^i$ with $r_i \in \mathbb{Z}_2, q_i \in \mathbb{Z}$
