Let \(C/F\) be an algebraic field extension. Prove that the following are equivalent:
- Every nonconstant polynomial in \(F[x]\) splits into linear factors over \(C\).
- Every algebraic extension \(E/F\) admits an \(F\)-embedding \(E\hookrightarrow C\).
Solution
1. Suppose every nonconstant polynomial in \(F[x]\) splits over \(C\).
First, \(C\) is algebraically closed. Indeed, let \(\beta\) be algebraic over \(C\). Since \(C/F\) is algebraic, \(\beta\) is algebraic over \(F\). Let \(m_{\beta,F}(x)\in F[x]\) be its minimal polynomial. By hypothesis this polynomial splits over \(C\). The minimal polynomial of \(\beta\) over \(C\) divides \(m_{\beta,F}\) in \(C[x]\), so it must be linear. Hence \(\beta\in C\).
Now let \(E/F\) be algebraic. Consider pairs \((K,\sigma)\) where \[ F\subseteq K\subseteq E \] and \(\sigma:K\hookrightarrow C\) is an \(F\)-embedding, ordered by extension. Every chain has an upper bound obtained by taking the union of the domains and embeddings, so Zorn’s lemma gives a maximal pair \((K,\sigma)\).
If \(K\ne E\), choose \(\alpha\in E\setminus K\). Let \(m_{\alpha,K}(x)\) be its minimal polynomial. Applying \(\sigma\) to the coefficients gives a polynomial over \(\sigma(K)\subseteq C\). Since \(C\) is algebraically closed, it has a root \(c\in C\). Sending \(\alpha\mapsto c\) extends \(\sigma\) to an embedding \(K(\alpha)\hookrightarrow C\), contradicting maximality. Hence \(K=E\).
2. Suppose every algebraic extension of \(F\) embeds in \(C\).
Let \(f\in F[x]\) be nonconstant and let \(E/F\) be a splitting field of \(f\). Then \(E/F\) is finite algebraic, so there is an \(F\)-embedding \[ \sigma:E\hookrightarrow C. \] All roots of \(f\) lie in \(E\), and their images under \(\sigma\) are again roots of \(f\) because \(\sigma\) fixes \(F\). Therefore every root of \(f\) lies in \(C\) up to this embedded copy of \(E\), and \(f\) splits into linear factors over \(C\).