Consider the map $\varphi:\mathbb{Z}\to \mathbb{Z}[X]/(2X-1)$ given by $\varphi(a)=a+(2X-1)$ for all $a\in \mathbb Z$.
We know that the induced map $\operatorname{Spec}\varphi:\operatorname{Spec} \mathbb{Z}[X]/(2X-1)\to \operatorname{Spec} \mathbb Z$ given by $P\mapsto \varphi^{-1}(P) $ is a continuous function. I want to chech if it is open or closed.
We also know that for any ideal $I$ of $\mathbb{Z}[X]/(2X-1)$, $\overline{\operatorname{Spec}\varphi(V(I))}=V(\varphi^{-1}(I))$. Thus I feel that the map $\operatorname{Spec}\varphi$ need not be closed in this case. But I could not show it. Any help is appreciated.
$\operatorname{Spec}\mathbb{Z}[X]/(2x-1)=\operatorname{Spec}\mathbb{Z}-\{2\}$ so $\varphi$ is an open immersion and not closed.
you can see this fact by constructing an isomorphism between $A=\mathbb{Z}[x]/(2x-1)$ and $\mathbb{Z_2}$ (localization of $\mathbb{Z}$ at the set {2,4,8,...}) or you can see it directly.
first you can see 2 is a unit in $A$. consider $f(x)$ is a polynomial of degree n then image of $2^n f$ in $A$ is equal to image of an integer $m$. so $(f)=(m)$ and $\phi(f)=m'$ if $m=2^k m'$ and $m\, $is odd. it is easy too see that $\phi$ gives an isomorphism between $$\operatorname{Spec}\mathbb{Z}[X]/(2x-1)=\operatorname{Spec}\mathbb{Z}-\{2\}$$