Let \(f,g\in\mathcal O(\mathbb C)\) have no common zeros. Show that \[ (f,g)=(1), \] i.e. there exist \(r,s\in\mathcal O(\mathbb C)\) such that \[ fr+gs=1. \]
Hint: split \(1/(fg)\) as \(F+G\), with the poles of \(F\) supported on the zeros of \(f\) and the poles of \(G\) supported on the zeros of \(g\).
Solution
1. At every zero \(a\) of \(f\), the meromorphic function\[ \frac1{f g} \]has a well-defined finite principal part.
Because \(f\) and \(g\) have no common zeros, \(g(a)\neq0\). Thus near \(a\) the only pole of \(1/(fg)\) comes from the zero of \(f\), and its order is finite. Hence its Laurent expansion at \(a\) has a finite negative-power tail, which is the desired principal part.
2. There exists a meromorphic function \(F\) whose poles are exactly amongthe zeros of \(f\) and whose principal part at each such zero agrees with theprincipal part of \(1/(fg)\).
1. If \(f\) has no zeros, take \(F=0\).
3. Define\[ G=\frac1{fg}-F. \]Then the only poles of \(G\) occur at zeros of \(g\).
At a zero of \(f\), the principal part of \(F\) was chosen to equal that of \(1/(fg)\), so the difference has a removable singularity there. Away from the zeros of \(f\) and \(g\), both terms are holomorphic. Thus the only remaining possible poles are the zeros of \(g\).
4. The functions\[ s=fF, \qquad r=gG \]are entire.
Every pole of \(F\) occurs at a zero of \(f\), and its principal part came from \(1/(fg)\). Since \(g\) is nonzero there, the pole order of \(F\) is at most the zero order of \(f\), so multiplication by \(f\) removes it. Hence \(fF\) is entire.
Likewise every pole of \(G\) occurs at a zero of \(g\). From \[ G=\frac1{fg}-F \] and holomorphicity of \(F\) at the zeros of \(g\), the pole order of \(G\) is at most the corresponding zero order of \(g\). Thus \(gG\) is entire.
5. These entire functions satisfy\[ fr+gs=1. \]
From 3, \[ \frac1{fg}=F+G. \] Multiplying by \(fg\) gives \[ 1=fgF+fgG =g(fF)+f(gG) =gs+fr. \] Therefore \((f,g)=(1)\) in \(\mathcal O(\mathbb C)\).