Let \(\phi: \mathbb{C}[x] \to F\) be a ring homomorphism where \(F\) is a field and \(\phi(1)\neq 0\).
Give an example where \(\phi\) is not onto.
If \(\phi\) is onto, show that \(F \cong \mathbb{C}\).
Solution
1. There is a non-surjective example.
Take \[ F=\mathbb C(x), \] the rational-function field, and let \[ \phi:\mathbb C[x]\hookrightarrow\mathbb C(x) \] be the natural inclusion. Then \(\phi(1)=1\neq0\). The map is not surjective, since \[ \frac1x\in\mathbb C(x) \] is not a polynomial and hence does not lie in the image.
2. For any such homomorphism, \(\phi(1)=1_F\) and the restriction of \(\phi\) to \(\mathbb C\) is injective.
Since \[ \phi(1)^2=\phi(1), \] the element \(\phi(1)\) is an idempotent in the field \(F\). The only idempotents in a field are \(0\) and \(1\), and the hypothesis excludes \(0\). Thus \[ \phi(1)=1_F. \] Consequently the restriction \[ \phi|_{\mathbb C}:\mathbb C\longrightarrow F \] is a nonzero ring homomorphism from a field. Its kernel is therefore the zero ideal, so it is injective.
3. If \(\phi\) is surjective, then \(F\cong\mathbb C\).
If \(\phi\) is onto, the first isomorphism theorem gives \[ F\cong \mathbb C[x]/\ker\phi. \] Since \(F\) is a field, \(\ker\phi\) is a maximal ideal of \(\mathbb C[x]\). Every ideal of the PID \(\mathbb C[x]\) is principal, and every irreducible polynomial over the algebraically closed field \(\mathbb C\) is linear. Hence every maximal ideal has the form \[ (x-a) \qquad(a\in\mathbb C). \] Therefore, for some \(a\in\mathbb C\), \[ F\cong\mathbb C[x]/(x-a)\cong\mathbb C, \] where the last isomorphism is evaluation at \(a\).