Finite morphism

Remarks

If $f$ is finite, then clauses 1 and 2 hold for every open affine $V \subseteq Y$. A morphism $\operatorname{Spec} A \to \operatorname{Spec} B$ is finite if and only if $B \to A$ is a finite morphism of rings. A finite morphism is affine and of finite type. Example: for a field $k$, the morphism $\mathbb{A}^1_k \to \mathbb{A}^1_k$ given by $k[x] \to k[t]$, $x \mapsto t^2$, is finite, since $k[t] = k[x] \cdot 1 + k[x] \cdot t$. Non-example: the open immersion $\operatorname{Spec} k[x, x^{-1}] \to \operatorname{Spec} k[x]$ satisfies clause 1 but not clause 2. If $k[x, x^{-1}]$ were a finite $k[x]$-module, then $x^{-1}$ would be integral over $k[x]$. Since $k[x]$ is an integrally closed domain, this would give $x^{-1} \in k[x]$, which is false.