I’m trying to prove that for an (EDIT: irreducible) variety of positive dimension (or, to simplify, one-dimensional) V⊆ˉKnV \subseteq \bar K^n, we have

trdeg(ˉK(V):ˉK)>0.\operatorname{trdeg}(\bar K(V) : \bar K) > 0.

(KK is any field, and ˉK\bar K is its algebraic closure.)

To this end, I am attempting to show that any non-constant rational function f=gh∈ˉK(V)f = \frac{g}{h} \in \bar K(V) has an infinite image (where it is defined): that is,

f[V∖V(h)]f[V \setminus V(h)]

is an infinite subset of ˉK\bar K. (This implies that for any p∈ˉK[t]p \in \bar K[t], if p(f)=0p(f) = 0 then p(f(x))=0p(f(x)) = 0 for every x∈V∖V(h)x \in V \setminus V(h), so then pp is zero at all the values in f[V∖V(h)]f[V \setminus V(h)], which is infinite, so then p=0∈ˉK[t]p = 0 \in \bar K[t]; thus ff is transcedental over ˉK\bar K.) I have managed to show that V∖V(h)V \setminus V(h) is infinite, but I am struggling to think of a nice proof that this implies that f[V∖V(h)]f[V \setminus V(h)] is likewise infinite. Intuitively, it should be true: f[V∖V(h)]f[V \setminus V(h)] cannot be finite, because polynomials are not discontinuous.

The only rigorous proof I can think of deals with continuity of ff over the domain V∖V(h)V \setminus V(h), but of course this requires a topology on both V∖V(h)V \setminus V(h) and ˉK\bar K, which might make sense for ˉK=C\bar K = \mathbb C, but not so much for KK a finite field. (At least, I’m less comfortable with that.)

Is there some algebraic-geometrical proof, then, that a non-constant rational function on a positive-dimensional irreducible variety takes on infinitely many values?

Even hints are ok. I would rather completely solve this on my own, but I’m definitely stuck.

EDIT: one other thought I had was to show that f[V∖V(h)]f[V \setminus V(h)] is cofinite in ˉK\bar K: there are only finitely many values that ff does not take on. But I don’t know for sure that this is actually true…

EDIT 2: yet another thought. Maybe ff defines a rational map (I don’t know if that is precisely the right object to consider) into ˉK\bar K, so then then the image of ff is a variety (somehow) in ˉK\bar K, but it has positive dimension as well? I don’t know, just throwing out ideas.

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1

Hint: This is false for a reducible variety of any dimension. For example the function xx defined on V((x−1)(x+1))⊆A2V((x-1)(x+1)) \subseteq \mathbb{A}^2 takes the values −1,1-1, 1.

– basket

2 days ago

What is your definition of dimension?

– A. S.

12 hours ago

@A.S. I consider the dimension of VV to be the maximal length of a chain of irreducible varieties V1,…,VnV_1, \ldots, V_n such that ∅⊊V1⊊…⊊Vn⊊V.\emptyset \subsetneq V_1 \subsetneq \ldots \subsetneq V_n \subsetneq V.

– feralin

7 mins ago

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