Hartshorne defines a morphism $f:X\rightarrow Y$ to be an finite morphism if "there exists a covering of $Y$ by open affine subsets $V_i=\operatorname{Spec}(B_i)$, such that for each $i$, $f^{-1}(V_i)$ is affine, equal to $\operatorname{Spec}(A_i)$, where $A_i$ is a $B_i$-algebra which is a finitely generated $B_i$-module". In the example immediately following this definition, he talks about schemes of "finite type over $k$", which I assume just means that the morphism $X\rightarrow \operatorname{Spec}(k)$ is finite.
What disturbs me in this definition is the the part about being each $A_i$ being finitely generated $B_i$-module, rather than, say, a finitely generated $B_i$-algebra. For example, if $X=\operatorname{Spec}( k[x])=\mathbb{A}^1$, then $X$ is obviously an affine scheme over $k$ but the morphism $X\rightarrow \operatorname{Spec}(k)$ is not of finite type over $k$ since $k[x]$ is not a finitely generated $k$-vector space (though it finitely generated as a $k$-algebra). Thus, according to this definition, $\mathbb{A}^1$ is not of finite type over $k$, which seems wrong to me. Am I misunderstanding something here, or is this a typo?
I already mentioned this in the comments, but for completeness I'll put it here as an answer too. I was confusing the definition of "finite morphism" with "morphism of finite type", where the latter is defined using finitely generated algebras. So, to say that $$ is a scheme of finite type over $$ then must mean that $X\rightarrow \operatorname{Spec}(k)$ is a morphism of finite type.