if the fibers of all points are finite then will the map be finite?

If $f:X\to Y$ is a regular map between affine varieties then we say $f$ is finite if $k[X]$ is integral over $k[Y]$. If $f$ is finite then fibers of all points are finite. I think the converse of this statement is false but I can't give a counter example. Can you give a counterexample?

• If $f:X\to Y$ is a quasi-finite finite type morphism of schemes, then $f$ is finite if and only if $f$ is proper. If your schemes happen to be affine, then $f$ is separated, so that $f$ is finite if and only if $f$ is universally closed. Mar 10 '16 at 8:29

It is not true that a map $f \colon X \to Y$ which is quasi-finite (i.e. has finite fibres) is finite - you need to also assume that $f$ is universally closed (see p.100 of Hartshorne).

To complement Derek's answer, here is another example where this fails: consider the map on spectra induced by the ring map $k[x] \hookrightarrow \frac{k[x,y]}{(xy-1)}$, for a field $k = \overline{k}$. Geometrically, we have the hyperbola $xy=1$ and we are projecting this onto the $x$-axis. The fibres either consist of one point or are empty (the fibre above $x=0$).

However, this map is not finite: indeed, a finite morphism is closed, but the image of this map is the complement of the point $x=0$ in $\mathbb{A}^1_k$, which is open.

An open immersion has finite fibres but is generally not finite. In fact I think an open immersion is finite if and only if it is also a closed immersion, hence an isomorphism onto a connected component.

• That is correct. If you're a monomorphism and proper then you're a closed embedding. Mar 9 '16 at 20:48
• @Alex Maybe over an algebraically closed field? For instance $\text{spec} \mathbb C$ is proper over $\text{spec} \mathbb R$ and the map is a monomorphism... Mar 9 '16 at 23:09
• That's not a monomorphism! Monomorphism is invariant under base change. Mar 10 '16 at 0:05
• @Alex Hmm!... :) Mar 10 '16 at 0:24

Exercise II.3.5 in Hartshorne tells you that you are correct:

"Show by example that a surjective, finite-type, quasi-finite morphism need not be finite,"

where $f:X \to Y$ is quasi-finite if $f^{-1}(y)$ is a finite set for every $y\in Y$.

For a specific example, the distinction you want to keep in mind is finitely generated as an algebra versus finitely generated as a module. The solution I've seen before (not mine) is the map $$\text{Spec}\left( k[t,t^{-1}]\oplus k[t,(t-1)^{-1}]\right) \to \text{Spec}\left(k[t]\right)$$ for your choice of field $k$.

• thanks. but $k[t,t^{-1}]+k[t,(t-1)^{-1}]$ is not an affine variety
– ali
Mar 9 '16 at 18:23