While studying for an Algebra Qualifying Exam I stumbled with an exercise that says
"Describe the ring $\mathbb{Z}[x]/(2x)$".
I've done similar cases, like $\mathbb{Z}[x]/(x)\cong\mathbb{Z}$, $\mathbb{Z}[x]/(2,x)\cong\mathbb{Z}_2$ but in this case I don't have the intuition of what must be the ring.
I think maybe there is no closed form for the ring.
Can someone provide me any hints or ideas? Thanks!
Let us call this ring $R$. Then $R$ is generated as a $\mathbb{Z}$-algebra by an element $\varepsilon$ which satisfies $2\varepsilon=0$.
So an element of $R$ has the form $a_0+a_1\varepsilon+a_2\varepsilon^2+\dots+a_n\varepsilon^n$, and $a_i$ are only well-defined modulo $2$ for $i>0$.
Let us define $S=\mathbb{Z}\oplus \varepsilon\mathbb{Z}/2\mathbb{Z}[\varepsilon]$. The elements of $S$ are of the form $(a,\varepsilon P)$ whith $P\in \mathbb{Z}/2\mathbb{Z}[\varepsilon]$, and the product is the natural one: $$ (a,\varepsilon P)\cdot (b,\varepsilon Q) = (ab,\varepsilon(\bar{a}Q+\bar{b}P+\varepsilon PQ)).$$ In general, if $A$ is a commutative ring and $I\subset A$ is an ideal, we can form the ring $\mathbb{Z}\oplus I$ the same way.
I claim that $R\simeq S$. Indeed, consider the map $f:\mathbb{Z}[x]\to S$ given by $f(a+xP) = (a,\varepsilon \bar{P})$ where $\bar{P}\in \mathbb{Z}/2\mathbb{Z}[\varepsilon]$ is the reduction of $P$ mod $2$. It is easy to check that $f$ is a ring morphism, by the definition of the product in $S$. Furthermore, it is clearly surjective.
It remains to see that the kernel of $f$ is exactly $(2x)$ to conclude that $f$ induces an isomorphism $R\to S$. Now let $Q=a+xP\in \mathbb{Z}[x]$. Then $f(Q)=0$ means that $a=0\in \mathbb{Z}$ and $\varepsilon \bar{P}=0\in \mathbb{Z}/2\mathbb{Z}[\varepsilon]$, which means that $\bar{P}=0$, so $P$ has even coefficients. Thus $f(Q)=0$ is equivalent to $Q=2x M$ for some $M\in \mathbb{Z}[x]$, which proves the claim about the kernel.