Recall that the set of meromorphic functions M(X) over a Riemann surface X is a field, under natural multiplication and addition of functions. Moreover, if f:Y→X is a holomorphic map, then f induces a field homomorphism.
f∗:M(X)g→M(Y)↦g∘f
In this post, we are going to present the relationship between fields of meromorphic functions, given such maps.
Lemma. Let f:X→Y be a nonconstant holomorphic map between Riemann surfaces. Then
f∗:M(Y)φ→M(X)↦φ∘f
defines a field extension. In other words, f∗M(Y)⊂M(X) is a subfield.
Proof. Consider 1Y:Y→C. Then f∗(1Y)=1Y∘φ=1X, which means that 1X∈f∗M(Y). Additionally, f∗M(Y) is closed under addition, multiplication and inverses. Thus f∗M(Y) is a subfield. □
Field extension actually gives another possible kinds of degree. Now, we have the following notion of degrees
- degree of a covering map;
- degree of a polynomial (affine plane curve);
- degree of a field extension.
The applicable fact is that: these three kinds of curves coincide.
Here comes the main theorem of this post.
Theorem. Let f:X→Y be a nonconstant holomorphic branched covering of degree d.
f∗:M(Y)↪M(X)
Set K=M(Y). Then
-
∀g∈M(X), the degree of field extension [K(g):K]≤d;
-
If [K(g):K]=d, then M(X)=K(g);
-
If g has d distinct values on f−1(b) for some b∈Y, then [K(g):K]=d.
Combining 2 and 3: if a meromorphic function g∈M(X) can be found, then the degree of field extension is d. In other words, [M(X):M(Y)]=d.
![Branched Covering](/img/branchCovering.png)
Proof. (1) Set S=R(f)∪{p∈X∣f(p)=∞} be the set of singular points in X, the subspace Y′=Y−f(S), and X′=X−S. Then by the theorem in Covering, f′:X′→Y′ is unramified.
⇒ ∀b∈Y′, ∃V⊂Y′ an open neighborhood of b, s.t.
f−1(V)=i=1⨆dUi,
where φi:V→Ui is a biholomorphic map. Let g∈M(X) and set
gi=g∘φi:V⊂Y′→C
Consider φ:M(Y)→C, a polynomial with indeterminant w∈M(V), defined by
φ(w)=i=1∏d(w−gi)=:i=0∑dbiwd−1
where bi∈O(V), and φ(f∗g)=0. The function bi∈M(V) can be extended to global meromorphic function Bi∈M(Y). Thus, φ∈M(Y)[w], and φ(f∗g)=0, which implies that φ is a polynomial that vanishes f∗g. Since degφ≤d, then
[K(g):K]≤d.
This proves the first statement.
(2) Suppose that [K(g):K]=d. Let h∈M(X) be another element. Then there exists a primitive element p∈M(X), such that
K⊂K(g)⊂K(g,h)=K(p)
Then
d≥[K(p):K]=[K(g,h):K(g)]⋅[K(g):K]=[K(g,h):K(g)]⋅d
which means that the equality must hold. In other words, [K(g,h):K(g)]=1. Therefore, h∈K(g) and we are done for the second statement.
(3) Suppose that g∈M(X) and b∈Y has distinct preimage under f. Let gi=g∘φi as before. If gi(b)=gj(b) for some i=j, then g cannot be a solution to a polynomial equation of degree d. THERE IS STILL A GAP HERE!?!