Let \(\mathcal O(\mathbb C)\) and \(\mathcal M(\mathbb C)\) be the rings of entire and meromorphic functions on \(\mathbb C\).
Show that \(\mathcal O(\mathbb C)\) is integrally closed in \(\mathcal M(\mathbb C)\); that is, every \(f\in\mathcal M(\mathbb C)\) satisfying a monic polynomial in \(\mathcal O(\mathbb C)[X]\) actually lies in \(\mathcal O(\mathbb C)\).
Is \(\mathcal M(\mathbb C)\) algebraically closed?
Solution
1. Let \(f\in\mathcal M(\mathbb C)\) satisfy a monic relation\[ f^d+a_{d-1}f^{d-1}+\cdots+a_1f+a_0=0, \qquad a_j\in\mathcal O(\mathbb C). \]Then \(f\) has no poles.
Suppose instead that \(f\) has a pole of order \(m>0\) at some point \(p\). Write \[ \operatorname{ord}_p(f)=-m. \] Then \[ \operatorname{ord}_p(f^d)=-dm. \] For every \(j<d\), holomorphicity of \(a_j\) gives \[ \operatorname{ord}_p(a_jf^j) =\operatorname{ord}_p(a_j)+j\operatorname{ord}_p(f) \ge-jm>-dm. \] Thus \(f^d\) is the unique term in the monic relation with the lowest order at \(p\). Such a principal part cannot be cancelled by the remaining terms, a contradiction. Hence no pole exists.
2. Therefore \(\mathcal O(\mathbb C)\) is integrally closed in\(\mathcal M(\mathbb C)\).
A meromorphic function on \(\mathbb C\) with no poles is holomorphic everywhere, hence entire. By 1, every meromorphic function integral over \(\mathcal O(\mathbb C)\) has no poles, so it belongs to \(\mathcal O(\mathbb C)\).
3. The field \(\mathcal M(\mathbb C)\) is not algebraically closed.
Consider \[ P(X)=X^2-z\in\mathcal M(\mathbb C)[X]. \] If it had a root \(h\in\mathcal M(\mathbb C)\), then \[ h^2=z. \] Taking orders at \(0\) would give \[ 2\operatorname{ord}_0(h) =\operatorname{ord}_0(z) =1. \] The left side is an even integer, which is impossible. Hence \(P\) has no root in \(\mathcal M(\mathbb C)\), so this field is not algebraically closed.