Let $k$ be a field, $V$ be an algebraic set. Suppose I have a regular finite map $\phi: V \rightarrow \mathbb{A}^n$. Here by finite I mean that $k[V]$ is a finite $k[x_1, ..., x_n]$-algebra.

I was wondering does this imply that $\phi$ is finite-to-one map? I would greatly appreciate explanation! Thank you very much!

  • 1
    $\begingroup$ Yes. Short answer: being finite is preserved by base change (in particular, computing fibers), and a finite algebra over a field has finitely many prime ideals (all of which are maximal). (A finite $k$-algebra is Artinian.) $\endgroup$ Jul 30 '16 at 16:10
  • $\begingroup$ @AreaMan I know you wrote "Short Answer", but would it be possible to give more explanation by any chance? as I am not really sure how it answers the question (because of my lack of background in algebraic geometry)... Thanks! $\endgroup$
    – Johnny T.
    Jul 30 '16 at 18:46
  • 1
    $\begingroup$ @Johny T. Okay I gave an answer. You don't really need the schemy stuff I remark about, and this can be done explicitly (its somewhere in Shaferevichs Projective Varieties book), but maybe it provides motivation for learning that language. $\endgroup$ Jul 31 '16 at 1:23

There are two basic facts at play here, the general formula for computing fiber products, and the fact that finite $k$-dimensional $k$-algebras correspond to finite spectra. Both of these are clarified by thinking about schemes as the "correct" objects, since otherwise you will leave the category of varieties, but if you know the spec functor it is sufficient - if you don't just keep in mind that for $A$ a finitely generated, reduced $k$-algebra, $Spec A$ "is" the variety with coordinate ring $A$.

About computing fibers:

If $f : X \to Y$ is a regular map, which induces a morphism $f^* : B \to A$ on the level of coordinate rings, then the fiber over a maximal ideal $m$ in $Y$ has coordinate ring $A \otimes_B B/m$. In other words, the ideal of functions that vanish on $f^{-1}([m])$ is generated by the pullbacks by $f$ of the functions that vanish at $[m]$ (the ideal $m$).

The categorical way to think about this is in terms of universal properties - you can verify that in the category of $B$-algebras $A \otimes_B B/m$ is the coproduct of $A$ and $B/m$, hence on the side of varieties the variety $\operatorname{Spec} A \otimes_B B/m$ is the product. (Really it is just finite type $k$-scheme in general, because it can have nilpotents.) Now, the fiber over a point should be the largest subvariety that maps down to that point, and this is what the universal property says if you unravel it. (I can say more if you like, but it is worth playing with this on your own.)

Geometrically, certainly the pullbacks of functions vanishing at $[m]$ vanish at $f^{-1}([m])$. For some geometric intuition, you can assume $V$ is affine and embed $V \subset A^m$, and assume that the point whose preimage you are considering is the origin in $A^n$... so that the fiber in $V$ above $[m]$ is given as the (scheme-theoretic) intersection of $V$ with some hypersurface cut out by $(f_1, \ldots, f_n)$, where these $f_i$ define $f : A^m \to A^n$. Then the $f_i$ cut out the fiber in $V$, and these are pulled back from the coordinate functions on $A^n$. (Behind the scenes here is that the scheme theoretic intersection is also given by a fiber product.)

Okay, so ... if you compute the coordinate ring of a fiber over $[m]$, which is $A := \phi^{-1}([m]) = k(V) \otimes_k L = k[x_1, \ldots, x_n] / f^*(m)$, where $k(V)$ the coordinate ring of $V$ as a $k$-algebra, you will get some finite $k$-dimensional algebra $A$ over $L = B/m$. (If $k$ is algebraically closed, then $k = L$).

(Note that $A$ need not be reduced - this means that it will not be a variety, but a scheme. As an example, consider the map $z \to z^2$ from $A^1$ to itself. Then the fiber above any point away from the origin is two points, but as you move towards the origin, these two points collide to make a non-reduced "fuzzy" point.)

We will show that the minimal primes of $A$ are also maximal. (That $A$ has minimal primes follows from $A$ being finite dimensional - either in the Krull sense or as a $k$ vector space.) Let $P$ be a minimal prime, and consider $A/P$. This is a domain, and a finite $k$-dimensional $k$-algebra. It is a good exercise to show that if finite $k$-dimensional domain is a field. Hence $P$ was maximal.

Moreover, by considering the chain $P_1 \supset P_1 \cap P_2 \supset \ldots$, and using that $A$ is artinian because it is finite dimensional, and using prime avoidance one can conclude that there are finitely many prime ideals in $A$. (Exercise.)

Hence the variety corresonding to $A / Nil(A)$ has finitely many points, or anyway $spec A$ is a finite set.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.