Prove that each field of characteristic zero contains a copy of the rational number field.
Solution
Let \(F\) be a field of characteristic zero, with multiplicative identity \(1_F\).
1. The map\[ \iota:\ZZ\longrightarrow F, \qquad \iota(n)=n\cdot1_F, \]is an injective ring homomorphism.
The ring-homomorphism identities follow from repeated addition. Its kernel is the ideal \[ \{n\in\ZZ:n\cdot1_F=0\}. \] Characteristic zero means this kernel is \(0\), hence \(\iota\) is injective.
2. The map extends uniquely to an injective field homomorphism\[ \widetilde\iota:\QQ\longrightarrow F, \qquad \widetilde\iota\!\left(\frac mn\right) =\iota(m)\iota(n)^{-1}. \]
If \(n\ne0\), injectivity of \(\iota\) gives \(\iota(n)\ne0\), so the displayed expression is defined. If \(m/n=m'/n'\), then \(mn'=m'n\), and applying \(\iota\) shows the two displayed values are equal, so the map is well-defined. The field operations are preserved by the usual common-denominator calculation. Finally, if \(\widetilde\iota(m/n)=0\), then \(\iota(m)=0\), hence \(m=0\) by 1; therefore the map is injective.
3. Hence \(F\) contains a subfield isomorphic to \(\QQ\).
The image of the injective field homomorphism in 2 is such a subfield.