Fields of meromorphic functions

Recall that the set of meromorphic functions M(X)\mathcal{M}(X) over a Riemann surface XX is a field, under natural multiplication and addition of functions. Moreover, if f:YXf:Y\rightarrow X is a holomorphic map, then ff induces a field homomorphism.

f:M(X)M(Y)ggf\begin{aligned} f^*:\mathcal{M}(X)&\rightarrow \mathcal{M}(Y)\\ g&\mapsto g\circ f \end{aligned}

In this post, we are going to present the relationship between fields of meromorphic functions, given such maps.

Lemma. Let f:XYf:X\rightarrow Y be a nonconstant holomorphic map between Riemann surfaces. Then

f:M(Y)M(X)φφf\begin{aligned} f^*:\mathcal{M}(Y)&\rightarrow \mathcal{M}(X)\\ \varphi&\mapsto\varphi\circ f \end{aligned}

defines a field extension. In other words, fM(Y)M(X)f^*\mathcal{M}(Y)\subset\mathcal{M}(X) is a subfield.

Proof. Consider 1Y:YC1_Y:Y\rightarrow\mathbb{C}. Then f(1Y)=1Yφ=1Xf^*(1_Y)=1_Y\circ\varphi=1_X, which means that 1XfM(Y)1_X\in f^*\mathcal{M}(Y). Additionally, fM(Y)f^*\mathcal{M}(Y) is closed under addition, multiplication and inverses. Thus fM(Y)f^*\mathcal{M}(Y) is a subfield. \square

Field extension actually gives another possible kinds of degree. Now, we have the following notion of degrees

  1. degree of a covering map;
  2. degree of a polynomial (affine plane curve);
  3. 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:XYf:X\rightarrow Y be a nonconstant holomorphic branched covering of degree dd.


Set K=M(Y)K=\mathcal{M}(Y). Then

  1. gM(X)\forall g\in \mathcal{M}(X), the degree of field extension [K(g):K]d[K(g):K]\leq d;

  2. If [K(g):K]=d[K(g):K]=d, then M(X)=K(g)\mathcal{M}(X)=K(g);

  3. If gg has dd distinct values on f1(b)f^{-1}(b) for some bYb\in Y, then [K(g):K]=d[K(g):K]=d.

Combining 2 and 3: if a meromorphic function gM(X)g\in\mathcal{M}(X) can be found, then the degree of field extension is dd. In other words, [M(X):M(Y)]=d[\mathcal{M}(X):\mathcal{M}(Y)]=d.

Branched Covering

Proof. (1) Set S=R(f){pXf(p)=}S=R(f)\cup\{p\in X|f(p)=\infty\} be the set of singular points in XX, the subspace Y=Yf(S)Y'=Y-f(S), and X=XSX'=X-S. Then by the theorem in Covering, f:XYf':X'\rightarrow Y' is unramified.

\Rightarrow bY\forall b\in Y', VY\exists V\subset Y' an open neighborhood of bb, s.t.

f1(V)=i=1dUi,f^{-1}(V)=\bigsqcup_{i=1}^d U_i,

where φi:VUi\varphi_i:V\rightarrow U_i is a biholomorphic map. Let gM(X)g\in\mathcal{M}(X) and set

gi=gφi:VYCg_i=g\circ\varphi_i:V\subset Y'\rightarrow\mathbb{C}

Consider φ:M(Y)C\varphi:\mathcal{M}(Y)\rightarrow\mathbb{C}, a polynomial with indeterminant wM(V)w\in\mathcal{M}(V), defined by

φ(w)=i=1d(wgi)=:i=0dbiwd1\varphi(w)=\prod_{i=1}^d(w-g_i)=:\sum_{i=0}^d b_iw^{d-1}

where biO(V)b_i\in\mathcal{O}(V), and φ(fg)=0\varphi(f^*g)=0. The function biM(V)b_i\in\mathcal{M}(V) can be extended to global meromorphic function BiM(Y)B_i\in\mathcal{M}(Y). Thus, φM(Y)[w]\varphi\in\mathcal{M}(Y)[w], and φ(fg)=0\varphi(f^*g)=0, which implies that φ\varphi is a polynomial that vanishes fgf^*g. Since degφd\deg\varphi\leq d, then

[K(g):K]d.[K(g):K]\leq d.

This proves the first statement.

(2) Suppose that [K(g):K]=d[K(g):K]=d. Let hM(X)h\in\mathcal{M}(X) be another element. Then there exists a primitive element pM(X)p\in\mathcal{M}(X), such that

KK(g)K(g,h)=K(p)K\subset K(g)\subset K(g,h)=K(p)


d[K(p):K]=[K(g,h):K(g)][K(g):K]=[K(g,h):K(g)]d\begin{aligned} d&\geq [K(p):K]\\ &=[K(g,h):K(g)]\cdot[K(g):K]\\ &=[K(g,h):K(g)]\cdot d \end{aligned}

which means that the equality must hold. In other words, [K(g,h):K(g)]=1[K(g,h):K(g)]=1. Therefore, hK(g)h\in K(g) and we are done for the second statement.

(3) Suppose that gM(X)g\in\mathcal{M}(X) and bYb\in Y has distinct preimage under ff. Let gi=gφig_i=g\circ\varphi_i as before. If gi(b)=gj(b)g_i(b)=g_j(b) for some iji\neq j, then gg cannot be a solution to a polynomial equation of degree dd. THERE IS STILL A GAP HERE!?!