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Let $X$ and $Y$ be schemes and $f:X\to Y$ a morphism between them.

We say that $f$ is of finite type if for every affine open $spec(A)=U\subset Y$ there exists a finite affine (open) cover of $f^{-1}(U)$ by $\{ V_i\}_{i=1}^n$, where $V_i=spec(B_i)$, such that $B_i$ is a finitely generated $A$-algbera.

On the other hand $f$ is finite if $f^{-1}(U)$ is affine, say $f^{-1}(U)=spec(B)$, and $A$ is finite $B$-module.

I was wondering why don't we have the intermediate notion as following.

The map $f$ is such that for every affine $U=spec(A)$ we have $f^{-1}(U)=\bigcup_{i=1}^n V_i$ where $V_i=spec(B_i)$ and $B_i$ is finite $A$-module. Let's call the this property quasi finite type.

So the question is as follows: Suppose $f$ a morphism of schemes was of quasi finite type then would it imply that it is actually finite.

If yes, then there is no need of such a definition.

If no, is it just because we don't see this kind of property in "nature" and hence don't need to define it.

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    Note that a localization of a finite module is no longer finite, so even if $B$ is a finite $A$ module and $spec B=spec B_0\cup spec B_1$ we dont have that $B_i$ is a finite $A$ module. – 2017-02-05
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    Your definition of quasi finite type will actually imply that the map is finite, so you do not need such a convoluted definition. – 2017-02-05
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    @Mohan could you please explain more as to why the map is finite. – 2017-02-05
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    I will give you a starting point. Let $f:X\to Y$ be a morphism of irreducible varieties, where $Y$ is affine. Let $U\subset X$ be an open affine set with $f:U\to Y$ be finite. Then $U=X$. Remember that finite morphisms are proper. – 2017-02-06

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The notion of a quasi-finite morphism already exists: it is a morphism of finite type with finite fibers.

Zariski's main theorem, in its Grothendieck's version, asserts that a quasi-finite morphism decomposes as an open immersion into a finite scheme.

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    I think the notion of quasi-finite you mentioned is different from the definition (of quasi finite type) in the question. – 2017-02-05
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    At the ring level, it is roughly a localisation $S_f$ of a finite $R$-algebra $S$. Isn't it the same? – 2017-02-05
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    I can't see how the definition in the question of quasi finite type implies finite fibers. – 2017-02-05
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    Finite morphisms have finite fibers. – 2017-02-05
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    But it is not a finite morphism, is it. – 2017-02-05
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    Your definition stipulates a covering by affine open sets corresponding to finite algebras. – 2017-02-05
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    From wiki page of quasi finite morphism an equivalent version tells me that for a morphism of finite type to quasi finite it is enough to have finite extension of residue fields. This is implied by my definition. So the map is indeed quasi finite. – 2017-02-05