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Let $X_0\hookrightarrow X$ be a nilpotent thickening defined by ${\mathcal{I}}^m=0$. If $X$ is an affine scheme, its closed subscheme $X_0$ is an affine scheme. Conversely, suppose $X_0$ is an affine scheme. The underlying topological spaces are identical, so $X$ is a quasi-compact scheme. For any quasi-coherent sheaf $\mathcal{F}$, each subquotient ${\mathcal{I}}^j\mathcal{F}/{\mathcal{I}}^{j+1}\mathcal{F}$ is naturally a quasi-coherent sheaf on $X_0$. By sheaf cohomology under a closed inclusion and vanishing of quasi-coherent cohomology on an affine scheme, its first sheaf cohomology on $X$ vanishes. The long exact sequence in sheaf cohomology and descending mathematical induction through this finite filtration prove $H^1(X,\mathcal{F})=0$. The Serre criterion for affineness now makes $X$ an affine scheme.

Take the dual numbers $A=k[\epsilon ]/({\epsilon }^2)$ over any field $k$, let $B=k$, and let $\alpha$ send $\epsilon$ to zero. Its kernel is the nonzero ideal $(\epsilon )$, but both spectra of rings have exactly one point: every prime ideal contains the nilpotent element $\epsilon$, and the only prime of $A$ is $(\epsilon )$. Consequently the induced map is a homeomorphism for the Zariski topology, although it is not an isomorphism of schemes:More generally, quotienting by a nilpotent ideal gives a nilpotent thickening with unchanged underlying topology. The structure sheaf retains information that the topology alone cannot recover.